{"id":301,"date":"2021-07-26T15:11:06","date_gmt":"2021-07-26T15:11:06","guid":{"rendered":"https:\/\/vk6gmd.com.au\/?page_id=301"},"modified":"2021-07-27T03:43:46","modified_gmt":"2021-07-27T03:43:46","slug":"mathematics","status":"publish","type":"page","link":"https:\/\/vk6gmd.com.au\/index.php\/mathematics\/","title":{"rendered":"Mathematics"},"content":{"rendered":"<p>[et_pb_section fb_built=&#8221;1&#8243; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; custom_padding=&#8221;||0px|||&#8221; global_module=&#8221;299&#8243; saved_tabs=&#8221;all&#8221;][et_pb_row _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221;][et_pb_column type=&#8221;4_4&#8243; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221;][et_pb_text _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; text_font=&#8221;Arial||||||||&#8221; text_font_size=&#8221;17px&#8221; min_height=&#8221;22.2px&#8221; custom_margin=&#8221;32px||-7px|||&#8221; custom_padding=&#8221;||0px|||&#8221; text_text_shadow_style=&#8221;preset2&#8243; text_text_shadow_blur_strength=&#8221;0.41em&#8221;]<\/p>\n<p><b><span lang=\"EN-US\">3<\/span><\/b><span lang=\"EN-US\">.<span>\u00a0<strong> Mathematics<\/strong><\/span><\/span><\/p>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row column_structure=&#8221;1_3,2_3&#8243; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; custom_padding=&#8221;||0px|||&#8221;][et_pb_column type=&#8221;1_3&#8243; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221;][et_pb_text _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221;]<\/p>\n<p>3.1\u00a0 \u00a0 \u00a0Mathematics<\/p>\n<p>[\/et_pb_text][et_pb_image src=&#8221;https:\/\/vk6gmd.com.au\/wp-content\/uploads\/2021\/07\/graph1.png&#8221; title_text=&#8221;graph1&#8243; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221;][\/et_pb_image][et_pb_image src=&#8221;https:\/\/vk6gmd.com.au\/wp-content\/uploads\/2021\/07\/graph2.png&#8221; title_text=&#8221;graph2&#8243; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221;][\/et_pb_image][\/et_pb_column][et_pb_column type=&#8221;2_3&#8243; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221;][et_pb_text _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; min_height=&#8221;34.6px&#8221; custom_padding=&#8221;||2px|||&#8221;]<\/p>\n<ul>\n<li><strong>Understand addition, subtraction, multiplication and division.<\/strong><\/li>\n<li><strong>Understand fractions, proper, improper and decimal. Understand powers of 10 (scientific notation).<\/strong><\/li>\n<li><strong>Recall units and sub-units; (tera, giga, kilo, UNIT, milli, micro, nano, pico) and conversion to and from each.<\/strong><\/li>\n<li><strong>Understand squaring and square roots. Understand reciprocals.<\/strong><\/li>\n<li><strong>Understand graphs and be able to interpret them.<\/strong><\/li>\n<\/ul>\n<p>[\/et_pb_text][et_pb_toggle title=&#8221;Addition and Subtraction&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p>In any maths problem it must be remembered that a number without a plus or minus sign written against it is assumed to be a positive number.\u00a0<\/p>\n<p>5 + 5 = 10\u00a0 is really (+5) + (+5) = (+10)<\/p>\n<p>Every number has a decimal point.\u00a0 A whole number such as 6 is six point zero (6\u22190), and for convenience sake we normally don\u2019t show the decimal zero.\u00a0 However it is just as correct to write 6\u22190000000000 as 6, the number is the same.<a name=\"_Toc252875397\"><\/a><\/p>\n<p>When <em>adding<\/em> numbers of the<em> same sign<\/em> the result will have the same sign as the numerals.\u00a0<\/p>\n<p>5+2 = 7\u00a0 or (+5) + (+2) = (+7)<\/p>\n<p>(-5) + (-2)\u00a0 =\u00a0 -7<\/p>\n<p>When <em>subtracting<\/em> numbers of the <em>same sign <\/em>the result will not always be as obvious. We are all familiar with\u00a0 five minus three equals two\u00a0 or (+5) \u2013 (+3) = (+2).<\/p>\n<p>(-5) \u2013 (-2) = (-3) may not be a major problem but<\/p>\n<p>(-2) \u2013 (-5) = (+3)\u00a0 could be confusing.<\/p>\n<p>It may be easier to treat all subtraction problems as additions <em>after working the signs.<\/em><\/p>\n<p>Like signs may be treated as a plus.\u00a0 That is to say \u2013 (-5) is the same as +5.\u00a0<\/p>\n<p>Using the above example :<\/p>\n<p>(-2) \u2013 (-5) the \u2013 (-5) being like signs become a + and we have<\/p>\n<p>(-2)\u00a0 + 5 and the answer becomes a little more obvious as +3<\/p>\n<p>Similarly (-5) \u2013 (-3) becomes (-5) + 3 = -2<\/p>\n<p>When adding values with <em>different signs<\/em> remember:<\/p>\n<ol>\n<li>The problem will always be done as a subtraction of the smaller number from the larger number.<\/li>\n<li>That the resultant sign will be the same as that of the highest value in the problem.<\/li>\n<li>When a number is shown without a sign it is assumed to be positive.<\/li>\n<\/ol>\n<p><strong>Examples:<\/strong><\/p>\n<ol>\n<li>(-5) + (+3)\u00a0 = -2<\/li>\n<\/ol>\n<p>Subtract the smaller number from the larger \u00a0[5-3 = 2]<\/p>\n<p>Apply the sign of the larger number\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [-2]<\/p>\n<ol start=\"2\">\n<li>(-3) + (+5) = 2<\/li>\n<\/ol>\n<p>Subtract the smaller number from larger \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [5-3 = 2]<\/p>\n<p>Apply the sign of the larger number\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [+2 or 2]<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;Fractions&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p><a name=\"_Toc252875398\"><\/a>The terms numerator and denominator are used to indicate a part of a fraction. <span style=\"font-size: 14px;\">The numerator is the part of the fraction above the fraction bar.<\/span><\/p>\n<p>The denominator is the part of the fraction below the fraction bar.<\/p>\n<p>1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Numerator<\/p>\n<p>\/ \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Fraction Bar<\/p>\n<p>2\u00a0\u00a0\u00a0\u00a0\u00a0 Denominator<\/p>\n<h2><a name=\"_Toc252875399\"><\/a><span>Proper Fractions<\/span><\/h2>\n<p>Proper Fractions are those which indicate a part of a whole.<\/p>\n<p><sup>1<\/sup>\/<sub>3<\/sub> indicates that the whole number is divided into 3 parts and only<\/p>\n<p>one part is present.<\/p>\n<p>It means the same as 1 divided by 3 and\u00a0 could be written as\u00a0 1\u00f73<\/p>\n<p>&nbsp;<\/p>\n<p>Often fractions can be simplified by dividing the numerator and denominator by the same number.<\/p>\n<p>For example:<\/p>\n<p><sup>3<\/sup>\/<sub>6<\/sub> is the same as <sup>1<\/sup>\/<sub>2<\/sub>\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 both denominator and numerator have been divided by 3.<\/p>\n<p><sup>8<\/sup>\/<sub>32<\/sub>\u00a0 =\u00a0 <sup>4<\/sup>\/<sub>16 <\/sub>\u00a0=\u00a0 <sup>2<\/sup>\/<sub>8<\/sub>\u00a0 =\u00a0 <sup>1<\/sup>\/<sub>4<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 both have been divided by two and two and two again.<\/p>\n<p><a name=\"_Toc252875400\"><\/a>Improper Fractions<\/p>\n<p>Improper fractions are those that indicate a complete number and parts.<\/p>\n<p><sup>4<\/sup>\/<sub>3 <\/sub>indicates that there is one whole (<sup>3<\/sup>\/<sub>3<\/sub>) and <sup>1<\/sup>\/<sub>3<\/sub>.\u00a0 This can be simplified to 1<sup>1<\/sup>\/<sub>3<\/sub><\/p>\n<p><sup>40<\/sup>\/<sub>32\u00a0\u00a0 <\/sub>=\u00a0\u00a0 <sup>20<\/sup>\/<sub>16\u00a0 <\/sub>\u00a0=\u00a0 <sup>10<\/sup>\/<sub>8\u00a0 <\/sub>\u00a0=\u00a0\u00a0 <sup>5<\/sup>\/<sub>4\u00a0 <\/sub>\u00a0=\u00a0 1<sup>1<\/sup>\/<sub>4 <\/sub>\u00a0\u00a0or<\/p>\n<p><sup>40<\/sup>\/<sub>32\u00a0 <\/sub>\u00a0=\u00a0 1 <sup>8<\/sup>\/<sub>32\u00a0 <\/sub>= 1<sup>4<\/sup>\/<sub>16\u00a0 <\/sub>=\u00a0 1<sup>2<\/sup>\/<sub>8 <\/sub>\u00a0=\u00a0 1<sup>1<\/sup>\/<sub>4<\/sub><\/p>\n<p><a name=\"_Toc252875401\"><\/a>Decimal Fractions<\/p>\n<p>A decimal is merely another way of expressing a fraction to the base ten.<\/p>\n<p>0\u22191 is 1 divided by ten, <sup>1<\/sup>\/<sub>10 <\/sub>\u00a0or one tenth.<\/p>\n<p>0\u22195 is 5 divided by ten, <sup>5<\/sup>\/<sub>10 <\/sub>or simplified to <sup>1<\/sup>\/<sub>2<\/sub>\u00a0 or one half.<\/p>\n<p>&nbsp;<\/p>\n<p>The number of decimal places indicates the number of zeros in the denominator.<\/p>\n<p>0\u22191 is <sup>1<\/sup>\/<sub>10 \u00a0\u00a0\u00a0<\/sub>:\u00a0\u00a0 0\u221901 is 1\/<sub>100\u00a0 <\/sub>\u00a0\u00a0:\u00a0\u00a0 0\u2219001 is <sup>1<\/sup>\/<sub>1000 <\/sub>\u00a0\u00a0\u00a0:\u00a0\u00a0\u00a0 0\u221900001 is <sup>1<\/sup>\/<sub>100000<\/sub>\u00a0 <sub><\/sub><\/p>\n<p>&nbsp;<\/p>\n<p>As with whole numbers, 0\u22195 could be written as 0\u221950 or 0\u2219500,<\/p>\n<p><strong>Examples:<\/strong><\/p>\n<p>0\u221950 is <sup>50<\/sup>\/<sub>100 <\/sub>which is the same as <sup>5<\/sup>\/<sub>10<\/sub> or 0\u22195<\/p>\n<p>0\u2219500 is <sup>500<\/sup>\/<sub>1000 <\/sub>which is the same as <sup>50<\/sup>\/<sub>100 <\/sub>\u00a0(or 0\u221950) or\u00a0 <sup>5<\/sup>\/<sub>10<\/sub> (or 0\u22195).<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;Powers of Ten&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p><span style=\"font-size: 14px;\">As we saw when squaring a number ten squared is one hundred (10<\/span><sup>2 <\/sup><span style=\"font-size: 14px;\">\u00a0= 100).<\/span><\/p>\n<p><a name=\"_Toc252875398\"><\/a>This can be expanded by using higher exponents such as 10<sup>3<\/sup> equals 1000<a name=\"_Toc252875404\"><\/a><\/p>\n<p>(10*10*10.\u00a0\u00a0\u00a0 10*10 = 100; 100*10 = 1000).<\/p>\n<p>10<sup>6<\/sup> = 1000000 ( [1] 10*10 = 100;\u00a0 [2] 100*10 = 1000;\u00a0 [3] 1000*10 = 10000;\u00a0<\/p>\n<p>[4] 10000*10 = 100000;\u00a0 [5]100000*10 = 1000000.<\/p>\n<p><strong>Note:<\/strong> \u00a0The exponent does <em>not<\/em> indicate the number of times the number is multiplied but the number of times the number is used.<\/p>\n<p>In 10<sup>3<\/sup> there are only two \u2018*\u2019 steps but the 10 is used three times and in the 10<sup>6<\/sup> there are five \u2018*\u2019 steps. the first step uses 10 twice (10*10) and each \u2018*\u2019 step after that uses 10 once so only five \u2018*\u2019 steps are used to multiply 10 by itself six times.\u00a0<\/p>\n<p>These exponents with a base 10 may be used with another number, such as 6&#215;10<sup>3<\/sup>.\u00a0 This is 6*10*10*10 or 6000.\u00a0<\/p>\n<p>(6*10 = 60; 60*10 = 600; 600*10 = 6000.\u00a0 Again the exponent shows how many times 10 is used in the multiplication process.\u00a0<\/p>\n<p>Exponents may also be used in the negative, such as 10<sup>-3<\/sup>.\u00a0 The minus sign in front of the exponent indicates that the number is divided by 10 to the power shown.\u00a0 In this case,<\/p>\n<p>divided by 10<sup>3<\/sup> or 1000, so 10<sup>-3<\/sup> is <sup>1<\/sup>\/<sub>1000<\/sub>.\u00a0 If combined with another number (5*10<sup>-3<\/sup>) it indicates the number (5) divided 1000 or <sup>5<\/sup>\/<sub>1000<\/sub>.<\/p>\n<p>Another way of writing <em>exponents with a base 10<\/em> is to use the letter E in place of the 10.<\/p>\n<p>5*10<sup>3<\/sup> could be written as 5E3 meaning five multiplied by ten to the power of three.<\/p>\n<p>Note the <em>E means a power of ten<\/em> so 2E2 is NOT 2<sup>2<\/sup> (4) but is 2*10<sup>2<\/sup> (200).\u00a0<\/p>\n<p>Convert 6 000 000 000\u00a0\u00a0\u00a0 (There are nine zeros = E9) = 6E9<\/p>\n<p>Convert 3E6\u00a0 (E6 indicates six zeros)\u00a0 = 3 000 000<\/p>\n<p>NOTE:\u00a0 In electronics we usually work with exponents in the multiple of 3 (E3, E6,E9 etc) because they equate to the SI Prefixes (to be explained shortly). but remember that exponents may be 2E2 (200) which is the same as 0\u22192E3 (200) or 20E1 (200).<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;S.I. Units&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p><a name=\"_Toc252875407\"><\/a>When using the decimal system any unit (meter, litre, gram etc) can be prefixed with an indicator to show larger or smaller amounts. Many of these are in common usage such as kilometre (1000m), millilitre (<sup>1<\/sup>\/<sub>1000<\/sub> ).\u00a0 The prefixes remain the same, regardless of the unit or what is being measured.\u00a0 Kilo can be applied to any measurement meaning a thousand of that unit whether it is a gram, meter, litre or snazzle. The only place where these are not used is in time measurement of greater than a second and less than a year.<\/p>\n<p>They can however be used to indicate time of less than a second.\u00a0\u00a0<\/p>\n<p>The prefixes go as high as Exa (1E18) and as small as atto (1E-18).\u00a0 In this course we are only concerned with units between Tera\u00a0 (1E12 and pico (1E-12).<\/p>\n<p>Prefix\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Number\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Power of ten<\/p>\n<p>Tera (T)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 1 000 000 000 000\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a01E12\u00a0\u00a0 (1x*10^12 or 1*10<sup>12<\/sup>)<\/p>\n<p>Giga (G)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 1 000 000 000\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 1E9\u00a0\u00a0\u00a0\u00a0 (1*10^9\u00a0 or\u00a0 1*10<sup>9<\/sup>)<\/p>\n<p>Mega (M)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 000 000\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 1E6\u00a0\u00a0\u00a0\u00a0 (1*10^6\u00a0 or\u00a0\u00a0 1*10<sup>6<\/sup>)<\/p>\n<p>Kilo\u00a0 (k)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 000\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 1E3\u00a0 \u00a0\u00a0\u00a0(1*10^3\u00a0 or\u00a0\u00a0\u00a0 1*10<sup>3<\/sup>)<\/p>\n<p>Unit\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 1<\/p>\n<p>Milli (m)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <sup>1<\/sup>\/<sub>1000 <\/sub>or 0\u2219001\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a01E-3\u00a0\u00a0\u00a0 (1*10^-3\u00a0 or\u00a0 1*10<sup>-3<\/sup>)<\/p>\n<p>Micro (\u03bc or u) \u00a0\u00a0\u00a0\u00a0\u00a0 <sup>1<\/sup>\/<sub>1 000 000 <\/sub>or 0\u2219000 001\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a01E-6\u00a0\u00a0\u00a0 (1*10^-6\u00a0 or\u00a0 1*10<sup>-6<\/sup>)<\/p>\n<p>Nano (n)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 <sup>1<\/sup>\/<sub>1 000 000 000<\/sub> or 0.000 000 001\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 1E-9\u00a0\u00a0\u00a0 (1*10^-9\u00a0 or\u00a0\u00a0 1*10<sup>-9<\/sup>)<\/p>\n<p>Pico\u00a0 (p)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <sup>1<\/sup>\/<sub>1 000 000 000 000 <\/sub>or\u00a0 0.000 000 000 001\u00a0 \u00a0 \u00a0 1E-12\u00a0 (1*10^-12 or\u00a0 1*10<sup>-12<\/sup>)<\/p>\n<p>When using these prefixes it is usually easier to work in decimals than fractions.<\/p>\n<p><a name=\"_Toc252875408\"><\/a><strong>Converting S.I. Units<\/strong><\/p>\n<p>To convert one unit to another it is a matter of moving the decimal place to the left or right.\u00a0 To convert 5000m to km the decimal place moves three places to the left leaving us with 5km.\u00a0 Starting with 5 0 0 0 \u2219 0m\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 5 0 0\u22190\u00a0\u00a0\u00a0 is one place<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 5 0\u22190\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 is two places<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 5\u22190\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 is three places<\/p>\n<p>If converting in the other direction, 5km to m the decimal is moved in the opposite direction so 5\u22190km becomes 5000\u22190m.\u00a0 The decimal point and the following zero are normally not written in so 5km = 5000m.\u00a0<\/p>\n<p>Once the units scale is known the rule becomes:<\/p>\n<p>If moving <strong>upward on the scale<\/strong> the unit (1) the decimal point moves to the <strong>left three places<\/strong> for each prefix.<\/p>\n<p>So if converting 7pF to nF.\u00a0 The scale goes pico, nano, micro, milli, 1, kilo etc so we are moving upward on the scale by one prefix so\u00a0 the decimal point moves to the left three places.\u00a0<\/p>\n<p>7\u22190; move one place = 0\u22197 ; the second place\u00a0 =\u00a0 0\u221907 ; third place = 0\u2219007nF<\/p>\n<p>Convert 600kV to MV.\u00a0 Moving upward on the scale by one prefix the decimal point moves to the left by three places.\u00a0 600; moving one place = 60\u22190;\u00a0 second place = 6\u221900; third place = 0\u2219600.\u00a0 600kV = 0\u22196MV<\/p>\n<p>Convert 3nF to mF.\u00a0 In this case we are moving through 2 prefixs (n, \u03bc, m) so the decimal moves six places to the left.\u00a0 3nF = 0\u2219000 003 mF.\u00a0<\/p>\n<p>If <strong>moving downward on the scale<\/strong> the decimal moves to the <strong>right <\/strong>three places for each prefix.<\/p>\n<p>Convert 4MV to V.\u00a0 Moving down the scale move by two prefixes move the decimal point six places to the right.\u00a0 One place = 40; two places = 400; three places = 4 000; four places = 40 000; five places = 400 000; six places = 4 000 000.\u00a0 4MV = 4 000 000V<\/p>\n<p>Convert 500\u03bcH to pH.\u00a0 Moving down on the scale by two prefixs \u2013 move the decimal 6 places to the right.\u00a0 One place = 5 000; two places = 50 000;\u00a0 three = 500 000; four = 5 000 000;\u00a0 five = 50 000 000 and six = 500 000 000.\u00a0 500\u03bcH =\u00a0 500 000 000pH<\/p>\n<p>Convert 6km to m.\u00a0 Moving down on the scale the decimal point moves to the right.\u00a0 One prefix so it moves three places.\u00a0 6km = 6 000m<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;Square of a Number&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p><a name=\"_Toc252875402\"><\/a>A square of a number is the number multiplied by itself and is indicated by a superscript figure 2 following the number (called the exponent). <span style=\"font-size: 14px;\">For example 3 squared would be written 3<\/span><sup>2<\/sup><span style=\"font-size: 14px;\">.\u00a0 Because putting in a superscript is not always possible it may be written as 3^2 (the ^ indicates following number(s) are in superscript).<\/span><\/p>\n<p>This indicates that the 3 is multiplied by itself that is 3*3 = 9.<\/p>\n<p>10<sup>2<\/sup> = 100 or 10*10<\/p>\n<p>Note:\u00a0 The number is <em>not<\/em> multiplied by the exponent but by itself.<\/p>\n<p><a name=\"_Toc252875403\"><\/a>Square Root of a Number<\/p>\n<p>The square root of a number is the opposite of the square of a number.\u00a0 That is to say the square root (written as \u221a ) of a number is the number which is multiplied by itself to give the original number.<\/p>\n<p>For example \u221a4 is 2 (because 2*2=4)\u00a0 or the \u221a9 = 3 (3*3=9).<\/p>\n<p>For small simple numbers these can be calculated from \u2018times tables\u2019 learnt in primary schools but for larger or more complicated numbers a calculator is required.<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;Operations with exponents&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p><strong style=\"font-size: 14px;\">Multiplication with Exponents<\/strong><a name=\"_Toc252875402\"><\/a><\/p>\n<p>When multiplying two numbers with exponents, the numbers are multiplied and the exponents are added.<\/p>\n<p>2E3*2E3\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 The numbers (2*2) are multiplied as normal (4)<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The exponents (E3 and E3) are added (E6)<\/p>\n<p>giving a result of\u00a0 4E6.<\/p>\n<p>This can be proved by converting the original problem to whole numbers and multiplying:<\/p>\n<p>2E3*2E3 = 2000*2000 = 4 000 000\u00a0 or\u00a0 4E6<\/p>\n<p><strong>Examples:<\/strong><\/p>\n<p>Resolve\u00a0 5E3*2E6\u00a0<\/p>\n<p>Multiply the numbers (5*2) = 10 and add the exponents (E3+E6) = E9 gives 10E9<\/p>\n<p>Resolve 3E6*2E-3<\/p>\n<p>Mulitply the numbers (3*2) = 6 and add the exponents (E6+E-3) = E3 gives 6E3<\/p>\n<p>Resolve 4E-6*2E3<\/p>\n<p>Multiply the numbers (4*2) = 8 and add the exponents E-6+E3 = E-3 gives 8E-3<\/p>\n<p><a name=\"_Toc252875406\"><\/a><strong>Division with Exponents<\/strong><\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 When dividing numbers with exponents, the numbers are divided and the exponents subtracted.<\/p>\n<p>4E9 \u00f7 2E6 or <sup>4E9<\/sup>\/<sub>2E6\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The numbers are divided (4\u00f72) = 2<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The exponents are subtracted (9-6) =3<\/p>\n<p>Result\u00a0 2E3<sub><\/sub><\/p>\n<p>Prove: 4 000 000 000 \u00f7 2 000 000 = 2000 or 2E3<\/p>\n<p>Resolve 4E-6\u00f72E-3\u00a0 The numbers are divided (4\u00f72)=2 and exponents are subtracted<\/p>\n<p>\u00a0(-6-(-3)) = -3.\u00a0 Giving 2E-3<\/p>\n<p>Resolve 6E-6\u00f72E-3.\u00a0 6\u00f72 = 3 and E-6 \u2013 E-3 = E-3 giving\u00a0 3E-3<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;Reciprocals&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p><a name=\"_Toc252875409\"><\/a>The reciprocal of a number is that number inverted.\u00a0 That is to say if we take any number as a fraction and turn it upside down.<\/p>\n<p>For example 2 as a fraction is <sup>2<\/sup>\/<sub>1 <\/sub>,\u00a0 turned upside down it becomes <sup>1<\/sup>\/<sub>2<\/sub> <sub>\u00a0<\/sub>.<\/p>\n<p>The reciprocal of two is one half or 0\u22195.<\/p>\n<p>Similarly the reciprocal of 0\u22195 is 2.\u00a0 0\u22195 as a fraction is one half (<sup>1<\/sup>\/<sub>2<\/sub>) when inverted<\/p>\n<p>becomes <sup>2<\/sup>\/<sub>1<\/sub> or 2\u00a0<\/p>\n<p>To put it another way the reciprocal of a number is one divided by that number.<\/p>\n<p>The reciprocal of 2 is 1\u00f72 or <sup>1<\/sup>\/<sub>2<\/sub>\u00a0 or 0\u22195.<\/p>\n<p>The reciprocal of 0\u22195 is 1\u00f70\u22195 or <sup>1<\/sup>\/<sub>0\u22195<\/sub> or 2<\/p>\n<p>The reciprocal of\u00a0 5 is <sup>1<\/sup>\/<sub>5<\/sub> or 0\u22192<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;Graphs &#8211; (see examples on the left)&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p>Graphs are merely pictorial representations of one function versus another.<\/p>\n<p>\u00a0[see conversion chart &#8211; frequency to wavelength on the left]<\/p>\n<p>The frequency represented on the vertical axis and the wavelength on the horizontal with the diagonal line showing the relationship intersection.<\/p>\n<p>To use this type of graph find the frequency required and trace horizontally to the relationship intersection line then vertically to the wavelength scale and read the result.\u00a0 To convert wavelength to frequency, find the wavelength on the bottom scale and trace up to the intersection line then horizontally to determine the frequency.<\/p>\n<p>A graph where voltage\/current is shown in relation to time. [see to the left]<\/p>\n<p>To translate this type of graph firstly the time interval markings must be known.\u00a0 If we assume they are 0\u22195 seconds and we wish to know the voltage at 2 seconds.<\/p>\n<p>Find the second point on the horizontal scale (fourth mark) and trace vertically until intersection with the sine wave and trace horizontally to read the vertical scale.\u00a0 The result is 1V.<\/p>\n<p>To find the time at which the voltage is 4V trace a line horizontally from the 4 on the vertical scale to intersection with the sine wave and drop vertically to read the time on the horizontal scale.\u00a0<\/p>\n<p>NOTE:\u00a0 There are several times when this occurs and unless further specification is given the answer will be 0\u22192,\u00a0 1\u22191, 2\u22197 and 3\u22196.\u00a0<\/p>\n<p>It is possible to predict occurrences using a graph such as this.<\/p>\n<p>For example: Calculating the time between existing 4v points. While the wave goes through its peak voltage there is 0\u22199 seconds between existing 4V points and while the wave goes through the minimum voltage there is 1\u22196 seconds between 4V points, so by adding the time through minimum voltage the next 4V peak should appear at 3\u22196 + 1.6 seconds and the one after will be at 3\u22196 + 1\u22196 + 0\u22199.\u00a0 They should occur at 5\u22192 secs and 6\u22191 secs.\u00a0<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;Formulae Transposition (optional)&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p>When dealing with a large number of formulae it is virtually impossible to remember all the variations of each formulae.\u00a0 It is essential to be able to transpose, or move the formulae around so as to obtain the desired result.<\/p>\n<p>If we take the Ohms law formulae (V=IR), we may be able to remember the three variations of it, V=IR; I=<sup>V<\/sup>\/<sub>R<\/sub> and R=<sup>V<\/sup>\/<sub>I<\/sub> but when we come to trying to remember all the variations of several long formulae, which are similar, it becomes much harder.\u00a0 It is necessary to be able to manipulate, or transpose formulae.\u00a0<\/p>\n<p>If we look at a maths problem:\u00a0 8 = <sup>16<\/sup>\/<sub>2 <\/sub>\u00a0it makes perfect sense.<\/p>\n<p>We can transpose that to read\u00a0 8*2 = 16.<\/p>\n<p>When using simple figures this will appear obvious, but what have we really done mathematically?<\/p>\n<p>We have multiplied both sides of the equation by two.\u00a0 On the left hand side 8*2, and on the right <sup>16<\/sup>\/<sub>2<\/sub> * <sup>2<\/sup>\/<sub>1<\/sub>\u00a0 =\u00a0 <sup>32<\/sup>\/<sub>2 <\/sub>= 16 ( another way of explaining it is that the 2 numerator cancels the 2 denominator leaving <sup>16<\/sup>\/<sub>1<\/sub> which is 16).<\/p>\n<p>As long as we do the same thing to both sides of the equation it will remain true.<\/p>\n<p>\u00a0Manipulating letters in the same way is the basis of transposing equations.<\/p>\n<p>It is important to remember that a letter in an equation is really a number, the reason we use a letter is because it hasn\u2019t been given a specific value yet but it represents a relationship with another value also represented by a letter.<\/p>\n<p>In Ohms law for example the relationship between the three letters is true regardless of the figures actually present or used.\u00a0<\/p>\n<p>Using Ohms law formulae, V = IR, (or V = I*R the multiplied sign is assumed to be there if nothing is written) we can transpose this into two other formulae.<\/p>\n<p>If we wish to make I the subject of the formulae, that is to say, we are given V and R and have to find I.<\/p>\n<p>V = IR\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0and we wish to leave I on its own (make it the subject), we must remove \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 the R from the right of the equation.\u00a0 If we divide both sides by R we get:<\/p>\n<p><sup>V<\/sup>\/<sub>R<\/sub> = <sup>IR<\/sup>\/<sub>R<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <sup>R<\/sup>\/<sub>R<\/sub> will equal one (any number divided by itself must equal 1)\u00a0 Leaving:<\/p>\n<p><sup>V<\/sup>\/<sub>R<\/sub> = I\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>V = IR\u00a0 and we wish to find R.\u00a0 Divide both sides by I (<sup>V<\/sup>\/<sub>I<\/sub> = <sup>IR<\/sup>\/<sub>I<\/sub>) leaving us with <sup>V<\/sup>\/<sub>I<\/sub> = R<\/p>\n<p>It does not matter whether we say <sup>V<\/sup>\/<sub>I <\/sub>=R or R=<sup>V<\/sup>\/<sub>I<\/sub>\u00a0 the result is the same.<\/p>\n<p>Another example:<\/p>\n<p>P=VI\u00a0\u00a0 Find I<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide both sides by V to leave I on its own<\/p>\n<p><sup>P<\/sup>\/<sub>V<\/sub> = <sup>VI<\/sup>\/<sub>V\u00a0\u00a0 <\/sub>or\u00a0\u00a0 <sup>P<\/sup>\/<sub>V<\/sub> = I<\/p>\n<p>Using a more complex formulae the process is the same:<\/p>\n<p>If\u00a0\u00a0 X = PfHt\u00a0 Find H.\u00a0 As before we must isolate H so both sides must be divided by P, f and t.\u00a0 This can be done in individual steps: \u2013<\/p>\n<ol>\n<li>Divide both sides by P ( <sup>X<\/sup>\/<sub>P <\/sub>= <sup>PfHt<\/sup>\/<sub>P <\/sub>)<\/li>\n<li>Remove the p divided by p (<sup>X<\/sup>\/<sub>P <\/sub>= fHt )<\/li>\n<li>Divide both sides by f \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (<sup> X<\/sup>\/<sub>Pf\u00a0 <\/sub>= <sup>fHt<\/sup>\/<sub>f<\/sub> )<\/li>\n<li>Remove the <sup>f<\/sup>\/<sub>f<\/sub> (<sup>X<\/sup>\/<sub>Pf<\/sub> = Ht)<\/li>\n<li>Divide both sides by t (<sup>X<\/sup>\/<sub>Pft <\/sub>= <sup>Ht<\/sup>\/<sub>t<\/sub>)<\/li>\n<li>Remove the <sup>t<\/sup>\/<sub>t<\/sub> (<sup>X<\/sup>\/<sub>Pft<\/sub> = H)<\/li>\n<\/ol>\n<p>The procedure is the same with any formulae.<\/p>\n<p>Another complex formulae:\u00a0<\/p>\n<p>Z = \u221a(R<sup>2<\/sup> + X<sup>2<\/sup>)\u00a0\u00a0 The brackets around the R<sup>2<\/sup>+X<sup>2<\/sup> indicate that the \u221a applies to all the objects within the brackets, this means that to resolve this the R and X must be squared and added before the square root can be applied.<\/p>\n<p>To substitute figures:\u00a0 If R=4 and X=4 then the equation should be solved as:<\/p>\n<p>Z = \u221a (4<sup>2 <\/sup>+4<sup>2<\/sup>)\u00a0\u00a0 =\u00a0\u00a0 \u221a (16+16)\u00a0\u00a0\u00a0 = \u221a32\u00a0 = 5\u221965<\/p>\n<p>When\u00a0 transposing parts of this formulae the procedure is reversed or the square root must be removed from the right of the formulae before anything can be done with the parts within the square root sign.<\/p>\n<p>A square root that is squared is the original number\u00a0\u00a0<\/p>\n<p>Mathematically\u00a0 \u221a4<sup>2<\/sup>\u00a0 =\u221a (4*4)\u00a0 =\u00a0 \u221a16\u00a0\u00a0 = 4.<\/p>\n<p>Also in this formulae ( Z=\u221a(R2+X2) we are adding not multiplying.<\/p>\n<p>To remove an addition the value must be subtracted whereas when multiplying it is divided.<\/p>\n<p>To put that mathematically:<\/p>\n<p>To transpose\u00a0 3*3 = 9 we divide by 3 on both sides giving 3 = <sup>9<\/sup>\/<sub>3<\/sub><\/p>\n<p>to transpose\u00a0 \u00a03+3 = 6 we subtract 3 on both sides giving\u00a0 3 = 6-3<\/p>\n<p>Z = \u221a(R<sup>2<\/sup> + X<sup>2<\/sup>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Find R<\/p>\n<p>Firstly remove the square root by squaring both sides\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z<sup>2 <\/sup>= (\u221a(R<sup>2<\/sup>+X<sup>2<\/sup>))<sup>2<\/sup><\/p>\n<p>On the left the square root is cancelled by the square\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z<sup>2 <\/sup>= R<sup>2<\/sup>+X<sup>2<\/sup><\/p>\n<p>Subtract X<sup>2<\/sup> from both sides\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z<sup>2<\/sup> \u2013 X<sup>2<\/sup> = R<sup>2<\/sup>+X<sup>2<\/sup> \u2212 X<sup>2<\/sup><\/p>\n<p>X<sup>2<\/sup> \u2212 X<sup>2<\/sup> cancel each other\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z<sup>2<\/sup> \u2212 X<sup>2<\/sup> = R<sup>2<\/sup><\/p>\n<p>Square root both sides to remove the square from the R\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a(Z<sup>2<\/sup>\u2212X<sup>2<\/sup>) = \u221aR<sup>2<\/sup><\/p>\n<p>The \u221a and <sup>2 <\/sup>on the right cancel each other\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a(Z<sup>2<\/sup> \u2212 X<sup>2<\/sup>) = R<\/p>\n<p>When doing any transposition of formulae it is a good check to substitute simple numbers at the start and end of the transposition to check that it has been done correctly.\u00a0<\/p>\n<p>As examples using the formulae we have transposed above:<\/p>\n<ol>\n<li>V=IR to\u00a0 I = <sup>V<\/sup>\/<sub>R<\/sub>\u00a0\u00a0\u00a0<\/li>\n<\/ol>\n<p>If we substitute simple figures in the first equation then using the same figures the second should hold true.\u00a0 If we say I and R both are 2 then V will be 4<\/p>\n<p>Then in the first equation\u00a0 (V = IR )\u00a0\u00a0 4 = 2*2<\/p>\n<p>In the transposition\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0( I = <sup>V<\/sup>\/<sub>R<\/sub> )\u00a0\u00a0 2 = <sup>4<\/sup>\/<sub>2<\/sub>\u00a0 giving 2=2<\/p>\n<p>Mathematically true so the transposition has been done correctly<\/p>\n<ol start=\"2\">\n<li>X = PfHt<\/li>\n<\/ol>\n<p>Again if we use 2 for the value of all the items on the right of the equation<\/p>\n<p>X = 2*2*2*2\u00a0\u00a0 or 16 = 2*2*2*2<\/p>\n<p>Using the transposed formulae of\u00a0 <sup>X<\/sup>\/<sub>Pft<\/sub> = H\u00a0\u00a0 If we substitute the figures we have <sup>16<\/sup>\/<sub>2*2*2 <\/sub>= 2<\/p>\n<p>2*2*2 = 8\u00a0 so we have <sup>16<\/sup>\/<sub>8<\/sub> = 2 giving 2=2<\/p>\n<p>Mathematically true so the transposition has been done correctly.<\/p>\n<ol start=\"3\">\n<li>Z = \u221a(R<sup>2<\/sup> + X<sup>2<\/sup>)<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0 Z = \u221a( 2<sup>2<\/sup> + 2<sup>2<\/sup> )\u00a0\u00a0\u00a0 =\u00a0\u00a0\u00a0 \u221a (4 + 4)\u00a0\u00a0\u00a0\u00a0 =\u00a0\u00a0 \u221a8\u00a0\u00a0\u00a0 = 2\u2219828<\/p>\n<p>Transposed\u00a0\u00a0 \u221a(Z<sup>2<\/sup> \u2212 X<sup>2<\/sup>) = R<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a(2\u2219828<sup>2<\/sup> \u2212 2<sup>2<\/sup>) = 2<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a(8 \u2212 4) = 2<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a4 = 2<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 = 2\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>Mathematically correct so the transposition has been done correctly.<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;Ratios (optional)&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p>A ratio represents a relationship between two quantities and is usually written as 2<strong>:<\/strong>1.\u00a0 This indicates that whatever value the left hand side has, the value on the right will always be half of that left value.\u00a0 The same ratio can be indicated by any pair of figures as long as the right is double the left, such as\u00a0 4<strong>:<\/strong>2; 8<strong>:<\/strong>4;\u00a0 20<strong>:<\/strong>10;\u00a0 400<strong>:<\/strong>200.\u00a0 The primary ratio in all these is 2:1.<\/p>\n<p>The important thing with ratios is to realize that the ratio can be considered as the total number of parts into which a whole divides.\u00a0 That is to say that if three is to be divided into a ratio of 2:1 the whole (three) is divided into three parts 2:1.\u00a0 Similarly five divided into a ratio of 2:1 is divided by three and distributed as\u00a0 <sup>2<\/sup>\/<sub>3<\/sub> : <sup>1<\/sup>\/<sub>3<\/sub>\u00a0 or\u00a0 3\u22193:1\u22196\u00a0<\/p>\n<p>18 divided into a ratio of 7:2<\/p>\n<p>7:2 is a total of 9 parts\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (7+2=9)<\/p>\n<p>18 divided by nine gives 2 per part<\/p>\n<p>The parts are proportioned as 7 parts\u00a0 (7*2 = 14)\u00a0<\/p>\n<p>and 2 parts (2*2 = 4)<\/p>\n<p>18 divided into a ratio of 7:2 is 14:4<\/p>\n<p>Divide 64 into a ratio of 5:3, resolves to <sup>5<\/sup>\/<sub>8 <\/sub>*64 : <sup>3<\/sup>\/<sub>8<\/sub>*64\u00a0 or\u00a0 40:24.<\/p>\n<p>Another problem involving ratios comes about when part of the ratio is given and the other part has to be found or the total value.\u00a0<\/p>\n<p>In a ratio of 3:2 the left side has a value of 15. What is the value of the other side and the total?<\/p>\n<p>In a ratio of 3:2 the total value is divided into five and proportioned as 3 parts to 2 parts or<\/p>\n<p>So we are told that the left side ( <sup>3<\/sup>\/<sub>5 <\/sub>of the total)\u00a0 has a value of 15.\u00a0 So <sup>1<\/sup>\/<sub>5<\/sub> will be 15\u00f73 or 5<\/p>\n<p>The right side is <sup>2<\/sup>\/<sub>5 <\/sub>\u00a0or 2*5 = 10<\/p>\n<p>The proof of this is that 15:10 reduces to 3:2 (both sides divided by 3).<\/p>\n<p>The total is obviously the right side plus the left or 10+15=25<\/p>\n<p>In a ratio of 7:2 the right hand side has a value of 6. What is the value of the left hand side and the total value?<\/p>\n<p>6=<sup>2<\/sup>\/<sub>9<\/sub> of the total therefore <sup>1<\/sup>\/<sub>9<\/sub> = 3 (6\u00f72) and <sup>7<\/sup>\/<sub>9<\/sub> =21 (3*7)<\/p>\n<p>The values are 21:6 (To prove &#8211; dividing both sides by 3 gives the original ratio &#8211; 7:2)<\/p>\n<p>The total value is the sum of both sides 21+6=27<\/p>\n<p>So far we have only worked ratios between two values but they may be expressed with any number of values.\u00a0 A ratio of\u00a0 5:3:2:4:12\u00a0 can be worked with in exactly the same way as the two value ratios above.<\/p>\n<p>If fifty two is to be divided into the above ratio, firstly calculate the number of parts in the ratio. 5+3+2+4+12 = 26.\u00a0 Fifty two divided by twenty six is two so each part is worth 2 so the 54 divides as 10:6:4:8:24.<\/p>\n<p>If given one of the ratio values say the second value (3) is six.\u00a0 <sup>3<\/sup>\/<sub>26<\/sub> = 6 then <sup>1<\/sup>\/<sub>26 <\/sub>= 2 so the ratio resolves as 10:6:4:8:24 and the total is 10+6+4+8+24 = 52<\/p>\n<p>[\/et_pb_toggle][et_pb_toggle title=&#8221;Directly Proportional and Inversely Proportional (optional)&#8221; _builder_version=&#8221;4.9.10&#8243; _module_preset=&#8221;default&#8221; title_font=&#8221;|700|||||||&#8221; title_font_size=&#8221;14px&#8221; custom_margin=&#8221;-35px|||||&#8221;]<\/p>\n<p>These are two terms which need to be understood in relation to formula.<\/p>\n<p>If we say that two items in a formula are directly proportional it indicates that if one item is increased (or decreased) the other will increase (or\u00a0 decrease) in a direct relationship.<\/p>\n<p>Example:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 I = <sup>V<\/sup>\/<sub>R<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 I is directly proportional to V<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 If we raise V by doubling it then I will double<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 It will not necessarily be the same amount of change but by the\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 same proportion.<\/p>\n<p>In figures:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 8 = <sup>16<\/sup>\/<sub>2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/sub>Raising V by 8 will not cause I to increase by 8 but by a direct \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 proportion.\u00a0 <br \/>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 8 is one half of 16 so I will increase by half its value or 4.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12 = <sup>24<\/sup>\/<sub>2<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16 has been increased by half its value (8) and so 8 increases by\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 half its value (4).\u00a0<\/p>\n<p>An inverse proportion is exactly the reverse.\u00a0 If one value decreases the other will increase by the same proportion.<\/p>\n<p>Example:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 I = <sup>V<\/sup>\/<sub>R<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 I is inversely proportional to R<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 If we increase R then I will reduce. Again by the same proportion.<\/p>\n<p>Figures;\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 8 = <sup>16<\/sup>\/<sub>2<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 If we increase 2 by doubling it to 4 the 8 will reduce by the same\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0proportion.\u00a0 The inverse of 2 is \u00bd so 8 will reduce by \u00bd to 4.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4 = <sup>16<\/sup>\/<sub>4<\/sub><\/p>\n<p>[\/et_pb_toggle][\/et_pb_column][\/et_pb_row][\/et_pb_section]<\/p>\n","protected":false},"excerpt":{"rendered":"<p>3.\u00a0 Mathematics3.1\u00a0 \u00a0 \u00a0Mathematics Understand addition, subtraction, multiplication and division. Understand fractions, proper, improper and decimal. Understand powers of 10 (scientific notation). Recall units and sub-units; (tera, giga, kilo, UNIT, milli, micro, nano, pico) and conversion to and from each. Understand squaring and square roots. Understand reciprocals. Understand graphs and be able to interpret them. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_et_pb_use_builder":"on","_et_pb_old_content":"","_et_gb_content_width":""},"_links":{"self":[{"href":"https:\/\/vk6gmd.com.au\/index.php\/wp-json\/wp\/v2\/pages\/301"}],"collection":[{"href":"https:\/\/vk6gmd.com.au\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/vk6gmd.com.au\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/vk6gmd.com.au\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/vk6gmd.com.au\/index.php\/wp-json\/wp\/v2\/comments?post=301"}],"version-history":[{"count":13,"href":"https:\/\/vk6gmd.com.au\/index.php\/wp-json\/wp\/v2\/pages\/301\/revisions"}],"predecessor-version":[{"id":404,"href":"https:\/\/vk6gmd.com.au\/index.php\/wp-json\/wp\/v2\/pages\/301\/revisions\/404"}],"wp:attachment":[{"href":"https:\/\/vk6gmd.com.au\/index.php\/wp-json\/wp\/v2\/media?parent=301"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}