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Flashcards by katiehumphrey, updated more than 1 year ago

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GCSE Math Flashcards on All math revision, created by katiehumphrey on 01/16/2014.

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Question | Answer |

direct proportion If two quantities are in direct proportion, as one increases, the other increases by the same percentage. If y is directly proportional to x, this can be written as y ∝ x | If y is directly proportional to x. When x = 12 then y = 3 Find the constant of proportionality and the value of x when y = 8. We know that y is proportional to x so y = kx We also know that when x = 12 then y = 3 To find the value of k substitute the values y = 3 and x = 12 into y = kx 3 = k × 12 So k = 3/12 = 1/4 To find the value of x , when y = 8 substitute y = 8 and k = 1/4 into y = kx 8 = (1/4) x So x = 32 when y = 8 |

indirect proportion Inverse proportion is when one value increases as the other value decreases. | y is inversely proportional to x. When y = 3, x = 12 . Find the constant of proportionality, and the value of x when y = 8. y ∝ 1/x y = k/x So xy = k Substitute the values x = 12 and y = 3 into xy = k 3 × 12 = 36 So k = 36 To find the value of x when y = 8, substitute k = 36 and y = 8 into xy = k 8x = 36 So x = 4.5 |

Straight line y ∝ x y = kx This equation gives us a straight line. The gradient of the line is k. | staright.gif (image/gif) |

Quadratic y ∝ x2 y = kx2 This equation gives us a curve. The larger the value of k, the steeper the graph. | quad_.gif (image/gif) |

Cubic y ∝ x3 y = kx3 | cubic.gif (image/gif) |

Square root y ∝square root of x y = ksquare root of x You have similar shaped curves for any powers between 0 and 1. Again, increasing k will make the graph steeper. | square.gif (image/gif) |

Inverse proportion y ∝ 1/x ² y = k/x ² Inverse proportion leads to curved graphs. | inverse.gif (image/gif) |

surds square root of ab = square root of a x square root of b square root of a x square root of a = a | Simplify square root of 12 over square root of 6 Answer square root of 12 over square root of 6 = square root of 12 over 6 (in general square root of a over square root of b = square root of a over b) square root of 2 |

equations with fractions | How do we solve the equation x/2 − 4 = 3? One way is to multiply everything by 2: x/2 − 4 = 3 x - 8 = 6 x = 14 Or, alternatively, add 4 and then multiply both sides by 2: x/2 − 4 = 3 x/2 = 7 x = 14 |

graph transformations f(x) + a [0/a] f(+ ) = left f(x- ) = right 2f(x) = y coordinates x 2 -f(x) = reflection of f(x) f(2x) = 1/2 x coordinate | graph_25.gif (image/gif) |

sin cos tan graphs | Untitled_a.png (image/png) |

factorising | ( x + 2 ) ( x + 5 ) = x2 + 7x + 10 ( x + 2 ) ( x + 3 ) = x2 + 5x + 6 ( x - 3 ) ( x - 5 ) = x 2 - 8x + 15 ( x + 6 ) ( x - 5 ) = x2 + x - 30 ( x - 6 ) ( x + 5 ) = x2 - x - 30 |

quadratic equations To solve a quadratic equation, the first step is to write it in the form: ax2 + bx + c = 0. Then factorise the equation | Solve the equation: x2 - 9x + 20 = 0 Solution First, factorise the quadratic equation x2- 9x + 20 = 0 Find two numbers which add up to 9 and multiply to give 20. These numbers are 4 and 5. (x - 4) (x - 5) = 0 Now find the value x so that when these brackets are multiplied together the answer is 0. This means either (x - 4) = 0 or (x - 5) = 0 So x = 4 or x = 5. You can check these answers by substitutuing 4 and 5 in to the equation: x2- 9x + 20 Substituting 4 gives: 42 - 9 × 4 + 20 = 16 - 36 + 20 = 0 Substituting 5 gives: 52 - 9 × 5 + 20 = 25 - 45 + 20 = 0 |

completing the square This is another way to solve a quadratic equation if the equation will not factorise. It is often convenient to write an algebraic expression as a square plus another term. The other term is found by dividing the coefficient of x by 2, and squaring it. | Rewrite x2 + 6x as a square plus another term. The coeffient of x is 6. Dividing 6 by 2 and squaring it gives 9. x2 + 6x = (x2 + 6x + 9) - 9 = (x + 3)2 - 9 |

quadratic formula Solve 2x2 - 5x - 6 = 0 Here a = 2, b = -5, c = -6 | quadequationshirev2.png (image/png) |

the quadratic formula | quadequationshirev1.png (image/png) |

area_scale_factor.png (image/png) | To find the area of the large triangle, you will need to first find the linear/length scale factor as the Step 1. In general to find the scale factor we divide the large quantity with the small quantity. Scale factor = BIG/SMALL =6/4 =1.5 Then Step 2 we would find the Area scale factor by squaring the linear scale factor; ASF = LSF2 =1.52 = 2.25 Step 3 we would multiple the area scale factor with the small area to find the area for the large triangle. Area = 12 x 2.25 = 27cm2 |

A taxi firm charges £0.50 per mile plus a fixed charge of £2.00. Write down a formula for the cost (C) of hiring this taxi to travel 'n' miles. | Solution It costs £2 + £0.50 to travel 1 mile. It costs £2 + 2 x £0.50 to travel 2 miles. It costs £2 + 3 x £0.50 to travel 3 miles. So travelling for 'n' miles will cost £2 + n x £0.50. The formula is C = £2 + (n x £0.50). |

Collecting like terms To simplify an expression, we collect like terms. Look at the expression: 4x + 5x -2 - 2x + 7 | To simplify: The x terms can be collected together to give 7x. The numbers can be collected together to give 5. So 4x + 5x -2 - 2x + 7 simplified is 7x + 5. |

(2x + 5) squared | (2x + 5) (2x + 5) = 4x squared + 20x + 25 |

multiplying out brackets Multiply out: 3(4x - 7) | First multiply: 3 × 4x = 12x Then multiply: 3 × -7 = - 21 Therefore: 3(4x - 7) = 12x - 21 |

multiplying out negatives Multiply out the expression: -3(2n - 8) | First, multiply -3 × 2n = - 6n Then multiply -3 × -8 = 24 -6n + 24 Therefore, -3(2n - 8) = - 6n + 24 |

multiplying out 2 brackets Multiply out these two brackets: (x + 4) (x + 3) | Multiply everything in the first bracket by the second bracket: x (x + 3) + 4 (x + 3) = x2 + 3x + 4x + 12 = x2 + 7x + 12 |

FOIL first outer inner last | (n + 4) (n + 4) F n squared O 4n I 4n L 16 n squared + 8n + 16 |

proof prove (3n + 1) squared - (3n - 1) squared is a multiple of 6 | (3n + 1) (3n + 1) = 9n squared + 6n + 1 (3n -1) (3n -1) = 9n squared - 6n + 1 = 12n divided by 6 = 2n its true |

Index notation: used to represent powers | a2 means a × a and here the index is 2 b3 means b × b × b and here the index is 3 c4 means c × c × c × c and here the index is 4 etc. When there is a number in front of the variable: 4d2 means 4 × d × d 2e3 means 2 × e × e × e |

index multiplying and dividing | p^3 × p^7 = p^10, and s^5 ÷ s^3 = s^2 |

solving equations | x - 6 + 6 = 9 + 6 x = 15 |

solving inverse equations Here is how to solve 2a + 3 = 7 using inverses. | 2a + 3 = 7 Undo the + 3 by subtracting 3 2a + 3 - 3 = 7 - 3 2a = 4 Then undo the multiply by 2 by dividing by 2 - again on both sides 2a ÷ 2 = 4 ÷ 2 The answer is: a = 2 |

unknowns on both sides of the equation Solve the equation 3b + 4 = b + 12, and find the value of b. | Answer First, you need to get all the b terms on the same side of the equation. Subtract b from both sides. 3b - b + 4 = 12 Then simplify. 2b + 4 = 12 Subtract 4 from both sides. 2b = 8 To find the value of b, divide both sides by 2. b = 4 |

equations with brackets Solve the equation: 3(b + 2) = 15 | 3(b + 2) = 15 Multiply out the brackets. Remember, everything inside the brackets gets multiplied by 3. 3 × b + 3 × 2 = 15 When you have multiplied out the brackets you get: 3b + 6 = 15 Next, undo the + 6. In other words, do the inverse and subtract 6 from both sides. 3b + 6 - 6 = 15 - 6 So 3b = 9 |

trial and improvement Find the answer to the equation x3 – 2x = 25 to one decimal place. | First we'll try: x = 3 33 - (2 x 3) 3 x 3 x 3 - (2 x 3) 27 - 6 = 21; too small even if the answer is to 1 decimal place, trial an answer with more than 1 decimal place |

simultaneous equations | Solve these simultaneous equations and find the values of x and y. Equation 1: 2x + y = 7 Equation 2: 3x - y = 8 Add the two equations to eliminate the ys: 2x + y = 7 3x - y = 8 ------------ 5x = 15 x = 3 Now you can put x = 3 in either of the equations. Substitute x = 3 into the equation 2x + y = 7: 6 + y = 7 y = 1 So the answers are x = 3 and y = 1 |

solving inequalities Solve the expression 3x - 7 < 8 | 3x - 7 < 8 Then add 7 to both sides, to cancel out the -7: 3x < 15 Next, simplify the inequality by dividing both sides by the number in front of x - in this case 3. x < 5 |

double inequalities Solve the inequality -4 < 2x - 6 < 12 | Add 6 to all parts: 2 < 2x < 18 Divide all parts by 2: 1 < x < 9 |

equations for a straight line graph can contain | an x term, a y term, and a number |

tables for graph plotting | images.jpg (image/jpg) |

On a graph, parallel lines have the same gradient. For example, y = 2x + 3 and y = 2x - 4 are parallel because they both have a gradient of 2. | download.jpg (image/jpg) |

the lines y = 2x + 3 and y = -1/2 x -1 cross at right angles | graph_11.gif (image/gif) |

y = mx + c m = gradient c = y intercept | download__1_.jpg (image/jpg) |

cubic graphs | y = x^3 |

Calculate the points and then plot the graph for the equation y = x3 - x + 8 | Untitled.png (image/png) |

reciprocal graphs Graphs of the form y = 1/x, 2/x, 3/x | Untitled__d.png (image/png) |

quadratic sequences If the difference between the terms changes, this is called a quadratic sequence. If you use the formula n^2 + n to make a sequence, it means that: When n = 1 you get 1^2 + 1 = 2 When n = 2 you get 2^2 + 2 = 6 When n = 3 you get 3^2 + 3 = 12 When n = 4 you get 4^2 + 4 = 20 - giving the sequence 2, 6, 12, 20. | quad_sequence.gif (image/gif) |

histograms frequency density = frequency / class width | download__2_.jpg (image/jpg) |

frequency tables finding the mean and median | mean = add all (midpoint x frequency) then divide by total frequency median = total frequency + 1 divided by 2 |

frequency polygons dot in midpoint | polygon.gif.pagespeed.ce.PCZfIa5BIx.gif (image/gif) |

median upper quartile lower quartile and interquartile range of frequency | median = frequency + 1 divided by 2 lower quartile = half median upper quartile = median + LQ interquartile range = UQ - LQ |

stem and leaf diagram | hd_crd2_dia_05a.gif (image/gif) |

Untitled_e.png (image/png) | graph_40.gif (image/gif) |

box plots | figure_86.gif (image/gif) |

figure_89.gif (image/gif) | The probability of three heads is: P (H H H) = 1/2 × 1/2 × 1/2 = 1/8 P (2 Heads and a Tail) = P (H H T) + P (H T H) + P (T H H) = 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 = 1/8 + 1/8 + 1/8 = 3/8 |

Amit is 12 years old. His brother, Arun, is 9. Their grandfather gives them £140, which is to be divided between them in the ratio of their ages. How much does each of them get? | The ratio of their ages is 12:9. We can simplify this. Dividing by 3 gives 4:3. So Amit gets 4 parts, and Arun gets 3. This means that the money has to be divided into 7 parts (4 + 3). £140 ÷ 7 = £20, so 1 part is £20. Amit gets 4 parts: 4 × £20 = £80. Arun gets 3 parts: 3 × £20 = £60. |

standard form | 15 000 000 = 1.5 × 10 000 000 This can be rewritten as: 1.5 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 1. 5 × 10^7 |

adding and subtracting standard form | 4.5 × 10^4 + 6.45 × 10^5 = 45,000 + 645,000 = 690,000 = 6.9 × 10^5 |

multiplying and dividing standard form | To multiply powers you add, eg, 10^5 × 10^3 = 10^8 To divide powers you subtract, eg, 10^5 ÷ 10^3 = 10^2 |

parts of a circle | download__3_.jpg (image/jpg) |

circumference of a circle | pi x diameter |

area of a circle | pi x (radius squared) |

area of a semi circle | pi x (radius squared) divided by 2 |

perimeter of a semi circle | (pi x radius) + (2 x radius) |

circle theorems | circleTheorems.gif (image/gif) |

angle at the centre of a circle | figure_32.gif (image/gif) |

angles in the same segment are equal | figure_34.gif (image/gif) |

angles in a semi circle are 90 degrees The angle at the centre (AOB) is twice the angle at the circumference (APB). As AOB is 180°, it follows that APB is 90° | figure_36.gif (image/gif) |

Opposite angles in a cyclic quadrilateral add up to 180° A cyclic quadrilateral is a quadrilateral whose vertices all touch the circumference of a circle. The opposite angles add up to 180o. In the cyclic quadrilateral below, angles A + C = 180o, and angles B + D = 180o | figure_38.gif (image/gif) |

The angle between the tangent and radius is 90° A tangent to a circle is a line which just touches the circle. | figure_41.gif (image/gif) |

difficult circle theorem | figure_45.gif (image/gif) |

quadrilateral has been divided into two triangles, so the interior angles add up to 2 × 180 = 360°. | pentagon has been divided into three triangles, so the interior angles add up to 3 × 180 = 540° |

interior angles of polygons | Interior angle of a regular polygon = sum of interior angles ÷ number of sides |

The volume of a pyramid is: 1/3 × base area × perpendicular height The surface area of a pyramid is the area of the base + the area of all the other sides. | download__4_.jpg (image/jpg) |

The volume of a cone is: 1/3πr^2h The curved surface area of a cone is: πrL (where L is the slant height). Note this doesn't include the area of the base, which is a circle, area πr^2. | shape_10.gif (image/gif) |

The volume of a sphere is 4/3πr^3 The surface area of a sphere is: 4πr^2 | shape_12.gif (image/gif) |

volume and area | length × length = area length × area = volume area ÷ length = length volume ÷ area = length volume ÷ length = area |

3 figure bearings | download__5_.jpg (image/jpg) |

Pythagoras | a^2 + b^2 = c^2 |

diag_cuboid.gif (image/gif) | find length of AF AC^2 = 6^2 + 2^2 AC = √40 AF^2 = AC^2 + CF^2 AF^2 = 40 + 3^2 AF = √49 AF = 7cm |

figure_132.gif (image/gif) | We are given angle A and side AB. AB is the hypotenuse. BC is opposite angle A. Therefore we use the formula: sin = opposite / hypotenuse sin 30 = BC / 7 Multiply both sides by 7 7 x sin30 = BC Using a calculator we get BC = 3.5cm |

figure_133.gif (image/gif) | The answer is 4.77cm (3 sf). Can you see that PQ is opposite the given angle and that QR is adjacent to it? Use the formula: tan = opposite / adjacent tan50 = PQ / 4 4 × tan50 = PQ PQ = 4.77 (3 sf) |

figure_134.gif (image/gif) | We are given angle Z XZ is adjacent to angle Z. YZ is the hypotenuse. Therefore we use the formula: cos = adjacent / hypotenuse cos 25 = 5 over YZ This time our unknown side (YZ) is the denominator. If we multiply both sides by YZ, we get: YZ x cos 25 = 5 over YZ x YZ YZ x cos25 = 5 Now we can divide both sides by cos25: YZ x cos 25 over cos 25 = 5 over 25 YZ = 5 over cos 25 Using a calculator: YZ = 5 ÷ 0.9063 = 5.52cm (3 sf) |

figure_136.gif (image/gif) | AC is adjacent to angle C. BC is the hypotenuse. So we use the formula: cos = adjacent / hypotenuse cosC = 5/7 To calculate C we need to find the inverse of cos (INV cos or SHIFT cos): C = Inv cos(5/7) Using a calculator: C = 44.4° |

the sine rule | furthertrigonometryhirev2_1.png (image/png) |

figure_140.gif (image/gif) | R = 25.4° Using the sine we can write: sinR over 4 = sin75 over 9 Multiplying both sides by 4, we get sinR = sin75 over 9 x 4 = 0.4293... SO R = inv sin(0.4293) R = 25.4° (1dp) |

the cosine rule | furthertrigonometryhirev3_2.png (image/png) |

ma09060.gif (image/gif) | a^2 = b^2 + c^2 - 2bc cos A Substitute the values into the formula. a^2 = 7^2 + 3^2 - 2 × 3 × 7 cos 35 a^2 = 58 - 42 cos 35 a^2 = 23.5956 a = 4.86cm |

area of a triangle | ABC = 1/2 ab sin C |

figure_142.gif (image/gif) | Area = 12.4cm^2 The area of the triangle is 1/2 × 5.2 × 7.1 x sin42 = 12.4cm^2 |

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