Linear Law Questions: Math Quiz!

1) Two variables, x, and y are related by the equation

. Write down the y-intercept of the graph of y against

.

The equation can be expressed as y = 3(1/x) + 5, which is of the form Y = mX + c where Y = y, X = (1/x)

Explanation

The equation can be expressed as y = 3(1/x) + 5, which is of the form Y = mX + c where Y = y, X = (1/x)

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2) Two variables, x and y are related by the equation

. Write down the gradient of the graph of xy against

.

The gradient of the graph of xy against x is -3. This means that for every unit increase in x, the value of xy decreases by 3. In other words, the graph has a negative slope of -3.

Explanation

The gradient of the graph of xy against x is -3. This means that for every unit increase in x, the value of xy decreases by 3. In other words, the graph has a negative slope of -3.

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3) Two variables, x and y are related by the equation

.

Write down the gradient of the graph of lg y against x to 3 significant figures.

Taking lg to both sides of the equation, we get
lg(y) = lg3 - xlg4 which is of the form Y = mX + c where Y = lg(y), X = x

Explanation

Taking lg to both sides of the equation, we get
lg(y) = lg3 - xlg4 which is of the form Y = mX + c where Y = lg(y), X = x

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4) Two variables, x and y are related by the equation

.

Write down the y-intercept of the graph of lg y against lg x to 3 significant figures.

Taking lg to both sides of the equation, we get
2lg(y) + 2lg(x) = lg5. Rearranging, we get lg(y) = -lg(x) + (lg5)/2 which is of the form Y = mX + c where Y = lg(y), X = lg(x)

Explanation

Taking lg to both sides of the equation, we get
2lg(y) + 2lg(x) = lg5. Rearranging, we get lg(y) = -lg(x) + (lg5)/2 which is of the form Y = mX + c where Y = lg(y), X = lg(x)

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5) Two positive variables, x and y are related by the equation

.

Write down the y-intercept of the graph

of against

.

Dividing the equation throughout by x, we get y^2 = 2/(x^2)-3 which is of the form Y = mX + c where Y = y^2, X = 1/(x^2)

Explanation

Dividing the equation throughout by x, we get y^2 = 2/(x^2)-3 which is of the form Y = mX + c where Y = y^2, X = 1/(x^2)

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6) Two positive variables, x and y are related by the equation

.

Write down the gradient of the graph of

against

.

Dividing both sides of the equation by xy, we get 1/y + 6/x = 2. Rearranging, we get 1/y = -6/x + 2 which is of the form Y = mX + c where Y = 1/y, X = 1/(x)

Explanation

Dividing both sides of the equation by xy, we get 1/y + 6/x = 2. Rearranging, we get 1/y = -6/x + 2 which is of the form Y = mX + c where Y = 1/y, X = 1/(x)

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7) Given the graph shown below, express y in terms of x.

Y = 0.5x^(0.5) + 4.5

Y = 0.5x + 4.5x^(0.5)

Y = 0.5x + 4.5

Y = 0.5x + 4.5x^2

Let Y = mX + c ---- (1)
m = (8-5)/(7-1) = 0.5
Subst (1,5) into (1), we get c = 4.5
Thus y/(x^0.5) = 0.5(x^0.5) + 4.5. Expressing y in terms of x, we get y = 0.5x + 4.5(x^0.5)

Explanation

Let Y = mX + c ---- (1)
m = (8-5)/(7-1) = 0.5
Subst (1,5) into (1), we get c = 4.5
Thus y/(x^0.5) = 0.5(x^0.5) + 4.5. Expressing y in terms of x, we get y = 0.5x + 4.5(x^0.5)

8) The diagram shows a straight line graph obtained by plotting xy against x. Find y in terms of x.

Y = -0.5x + 3

Y = -0.5 + 3x

Y = -0.5x + 13

Y = -0.5 + (3/x)

Let Y = mX + c ---- (1)
m = (8-2)/(-10-2) = -0.5
Subst (2,2) into (1), we get c = 3
Thus xy = -0.5x + 3. Expressing y in terms of x, we get y = -0.5 + 3/x

Explanation

Let Y = mX + c ---- (1)
m = (8-2)/(-10-2) = -0.5
Subst (2,2) into (1), we get c = 3
Thus xy = -0.5x + 3. Expressing y in terms of x, we get y = -0.5 + 3/x

9) The variables x and y are related in such a way that when (y - x) is plotted against x², a straight line which passes (1,1) and (5, 3) is obtained. Find the values of x when y = 5.

X = 2.16, x = -4.16

X = -2.16, x = 4.16

X = -7.19, x =-2.19

X = 7.19, x = 2.19

Let Y = mX + c ---- (1)
m = (3-1)/(5-1) = 0.5
Subst (1,1) into (1), we get c = 0.5
Thus y-x = 0.5(x^2) + 0.5
When y = 5, 5-x = 0.5x^2+0.5. Simplifying, we get 0.5x^2+x-4.5 = 0. Solve for x to get x = 2.16, x = -4.16

Explanation

Let Y = mX + c ---- (1)
m = (3-1)/(5-1) = 0.5
Subst (1,1) into (1), we get c = 0.5
Thus y-x = 0.5(x^2) + 0.5
When y = 5, 5-x = 0.5x^2+0.5. Simplifying, we get 0.5x^2+x-4.5 = 0. Solve for x to get x = 2.16, x = -4.16

10) It is known that x and y are related by the equation ay = bx³ - x, where a and b are unknown constants. Express this equation in a form suitable for drawing a straight-line graph, and state which variable should be used for each axis. Which of the following statement is correct?

Plot (y/x) against (x^2), gradient = (b/a) and y-intercept = -(1/a)

Plot (y/x) against (x^2), gradient = (b/a) and (y/x)-intercept = -(1/a)

Plot (y/x) against (x), gradient = (-b/a) and y-intercept = -(1/a)

Plot (y/x) against (x), gradient = (-b/a) and (y/x)-intercept = -(1/a)

Divide the given equation by ax gives y/x = (b/a)x^2 - 1/a, which is of the form Y = mX + c where Y = y/x, X = x^2. When referring to the Y-intercept of this graph, we should call it the (y/x)-intercept and not y-intercept.

Explanation

Divide the given equation by ax gives y/x = (b/a)x^2 - 1/a, which is of the form Y = mX + c where Y = y/x, X = x^2. When referring to the Y-intercept of this graph, we should call it the (y/x)-intercept and not y-intercept.