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2 years ago

Ariel receives

Mathematics

2 answers:

tatyana61 [14]2 years ago

5 0

She can get $12. X= ($42)(2/7)

Anastaziya [24]2 years ago

4 0

Mark Brainliest please Answe $12

Step by step

1 day 2/7day
-------- = ----------
$42 $x

Cross multiply:

$x = ($42)*(3/7)

Solve for x.

x= $12

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2 years ago

Find the solution set of this inequality|10x+20| ≤10

Pavlova-9 [17]

Answer:

solution is

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Step-by-step explanation:

we are given

$|10x+20|\leq 10$

Firstly, we will find critical values

so, let's assume it is equal

$|10x+20|= 10$

now, we can break absolute sign

For $|10x+20|= -(10x+20)$ :

$-(10x+20)= 10$

we can solve for x

$-10x-20= 10$

Add both sides by 20

$-10x-20+20= 10+20$

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Divide both sides by -10

and we get

$x=-3$

For $|10x+20|= (10x+20)$ :

$(10x+20)= 10$

we can solve for x

$10x+20= 10$

Subtract both sides by 20

$10x+20-20= 10-20$

$10x= -10$

Divide both sides by 10

and we get

$x=-1$

so, critical values are

$x=-3$

$x=-1$

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

For $(-\infty,-3)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

$|10\times -5+20|\leq 10$

$|-50+20|\leq 10$

$30\leq 10$

so, this is FALSE

For $[-3,-1]$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

$|10\times -2+20|\leq 10$

$|-20+20|\leq 10$

$0\leq 10$

so, this is TRUE

For $(-1,\infty)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=0

$|10\times 0+20|\leq 10$

$|0+20|\leq 10$

$20\leq 10$

so, this is FALSE

so, solution is

[-3,-1]

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3 years ago