Linear functions: answer key for math exercises in 3rd grade.

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The answer key to the math exercises in 3ème on linear functions.

Calculate images and priors. To work on the curve of a linear function and to exploit a table of values.

Exercise 1:
Let be the linear function f : x 1,2x.

a. Calculate f(5); f(- 1,2); f(0); f(100).

f(5)=1,2×5=6 ; f(-1,2)=- 1,44 ; f(0)= 0 ; f(100 ) = 120

b. Compute the numbers x whose images are 2400 ; – 45.

$x=\frac{2400}{1,2}=200$

$x=\frac{-45}{1,2}=-37,5$

Exercise 2:
Let g be the linear function such that g 😡 – 0.4x.

a. What is the coefficient of the function g?

a = – 0,4

b. Calculate the images of 10; – 5 and 1.

g(10) = – 4; g(-5)=2; g(1) = -0.4

Exercise 3:
We know that 18 has the image of 23 by the function f and that 12 has the image of 14 by f.

Is f a linear function? why?

Let’s compare $\frac{23}{18}$ and $\frac{14}{12}$ :

$23\times \,12=276\,\,14\times \,18=252$

The fractions are not equal so it is not a linear function.

Exercise 4:

Express the linear function f in the form $x\mapsto \,\,\,ax$ ( the number a is to be determined), then calculate f(0); f(1) and f( – 2).

1. When the image of 10 is – 3.

We have $a=-\frac{3}{10}=-0,3$ so f is such that .

2. When f (- 100)= – 46.
We have $a=\frac{-46}{-100}=0,46$ so f is such that .

3. When the coefficient of f is 2.5.
We have f which is defined by .

Exercise 5:

In a reference frame, it is enough to know the coordinates of a single point ( whose abscissa is non-zero) belonging to the curve of a linear function to draw it.

a. For $f:x\,\mapsto \,\,\,2,5x$ .

We have f(2)=2.5×2 = 5 so the curve passes through the point A(2;5).

b. Draw the line d of equation y = 1.2x, we will note this function h.
We have h(5)=1,2×5=6 so the curve passes through the point B ( 5 ; 6).

c. Draw the line d’ representing the linear function g of coefficient a = – 2.

We have g(1)= – 2 so the curve passes through C(1;- 2).

Here are the curves of these three linear functions:

Exercise 10:
Here are the curves of these five linear functions.

y=2.5x is (d1)

y= – 3x is (d5)

$y=-\frac{1}{3}x$ ets (d3)

$y=\frac{1}{3}x$ is (d2)

$y=-\frac{5}{4}x$ is (d4)

Exercise 14:

f is a linear function. Determine the expression of f(x)

$f(\frac{12}{5})=\frac{8}{5}$

f is of type .

$f(\frac{12}{5})=a\times \,\frac{12}{5}=\frac{8}{5}$

$a=\frac{\frac{8}{5}}{\frac{12}{5}}=\frac{8}{5}\times \frac{5}{12}=\frac{8:4}{12:4}\,=\frac{2}{3}$

Conclusion: ${\color{DarkRed}\,f(x)=\frac{2}{3}x}$

Exercise 15:

a) Let: x be the initial price of an item and: y its final price after an increase or decrease. What is the percentage increase or decrease in each of the following cases? (1): y = 1.4x

$1,4=1+0,4=1+\frac{40}{100}$

That’s a 40% increase.

(2): y = 0.5x

$0,5=1-0,5=1-\frac{50}{100}$

That’s a 50% reduction.

(3): y = 0.9x

$0,9=1-0,1=1-\frac{10}{100}$

That’s a 10% reduction.

(4): y = 1.05x

$1,05=1+0,05=1+\frac{5}{100}$

That’s a 5% increase.

Exercise 16:

1. An object A costs 65 euros. Its price increases by 5%.

How much does it cost after this increase?

The multiplying factor is $k=1+\frac{5}{100}=1,05$ .

$65\times \,1,05=68,25$

The final price is 68.25 euros.

2. object B costs 88 euros after a 10% increase.

What was its price before this increase?

The multiplying factor is $k=1+\frac{10}{100}=1,1$ .

$\frac{88}{1,1}=80$

The price before increase is 80 euros.

3. an object C costs 45 euros. After an increase its price is 50,40 euros.

What is the percentage of this increase?

The multiplying factor is $k=\frac{50,40}{45}=1,12=1+0,12=1+\frac{12}{100}$

The increase was 12%.

Exercise 17:

A clothing store manager decides to lower his prices by 15%.

a) What is the linear function modeling this decrease?

The linear function modeling a 15% decline is f(x)=0.85x .

b) What is the new price of a pair of pants that cost 70 e before this decrease?

f(70)=0.85×70=59.5 €.

c) What is the old price of a sweater that costs 50,12 € after this decrease?

We look for x such that f(x)=50.12

where 0.85x=50.12

$x=\frac{50,12}{0,85}$

Conclusion: the old price of the sweater was 56 € .

Exercise 18:

Consider the linear function f with coefficient – 5.

So we have f(x)= – 5x

Calculate the image by f of the following numbers:

a) 0

f(0)=0 (f is a linear function).

b) 3

$f(3)=-5\times \,3=-15$

c) – 2

$f(-2)=-5\times \,(-2)=10$

d) $\frac{3}{7}$

$f(\frac{3}{7})=-5\times \,(\frac{3}{7})=-\frac{15}{7}$

e) $-\sqrt{3}$

$f(-\sqrt{3})=-5\times \,(-\sqrt{3})=5\sqrt{3}$

Exercise 19:

For each function, specify whether it is linear and, if so, its coefficient.

a) $f:x\,\mapsto \,3,5x$

It is linear with coefficient a= 3,5 .

b) $g:x\,\mapsto \,2+x$

it is not linear, it is an affine function.

c) $h:x\,\mapsto \,7x^2$

it is not linear, it is a square function.

d) $i:x\,\mapsto \,-x$

it is linear with coefficient a= – 1 .

e) $j:x\,\mapsto \,5$

it is not linear, it is a constant function.

f) $k:x\,\mapsto \,\frac{5x}{3}$

it is linear with coefficient $a=\frac{5}{3}$ .

Exercise 20:

A product of factors is zero if and only if at least one of the factors is zero.

$x=0\,\,ou\,\,x-4=0$

$x=0\,\,ou\,\,x=4$

Conclusion : The numbers 0 and 4 have the same image by V1 and V2 .

Exercise 21:

Let f be the linear function defined by: $f:\,x\,\mapsto \,-2x$ .

1. Calculate f(3), f( – 2), f(7).

$f(3)=-2\times \,3=-6$

$f(-2)=-2\times \,(-2)=4$

$f(7)=-2\times \,7=-14$

2. What are the images by f of – 1, 6, $\frac{3}{2}$ ?

$f(-1)=-2\times \,(-1)=2$

$f(6)=-2\times \,6=-12$

$f(\frac{3}{2})=-2\times \,\frac{3}{2}=-3$

3. Find the number whose image is 7.

$x=-\frac{7}{2}$

${\color{DarkRed}\,x=-3,5}$

The answer key to the math exercises on linear functions in 3rd grade.

After having consulted the answers to these exercises on linear functions in 3rd grade, you can return to the exercises in 3rd grade.

Third grade exercises.

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