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4 years ago

15

Jasmine wants to use her savings of $1,128 to buy video games and music cds. The total price of the music cd she bought was $72.

The video games cost $43 each. What is the maximum number of video games that jasmine can buy with her savings?

2 answers:

Firlakuza [10]4 years ago

8 0

24 video games.

First you subtract $72 from $1,128 then you divide $43 from $1,056 and you get 24.5581395
So the most video games jasmine can buy is 24.

Colt1911 [192]4 years ago

4 0

She can buy 24 video games plus the cd

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2 years ago

Which values are in the solution set of the compound inequality? Check all that apply. 4(x + 3) ≤ 0 or x + 1 > 3 –6 –3 0 3 8

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we have

$4(x + 3) \leq 0$ -------> inequality 1

or

$x + 1 > 3$ -------> inequality 2

we know that

In this system of inequalities, for a value to be the solution of the system, it is enough that it satisfies at least one of the two inequalities.

let's check each of the values

<u>case 1)</u> x=-6

<u>Substitute the value of x=-6 in the inequality 1</u>

$4(-6 + 3) \leq 0$

$4(-3) \leq 0$

$-12 \leq 0$ -------> is ok

The value of x=-6 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

<u>case 2)</u> x=-3

<u>Substitute the value of x=-3 in the inequality 1</u>

$4(-3 + 3) \leq 0$

$4(0) \leq 0$

$0 \leq 0$ -------> is ok

The value of x=-3 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

<u>case 3)</u> x=0

<u>Substitute the value of x=0 in the inequality 1</u>

$4(0 + 3) \leq 0$

$4(3) \leq 0$

$12 \leq 0$ -------> is not ok

<u>Substitute the value of x=0 in the inequality 2</u>

$0 + 1 > 3$

$1 > 3$ --------> is not ok

The value of x=0 is not a solution of the compound inequality

case 4) x=3

<u>Substitute the value of x=3 in the inequality 1</u>

$4(3 + 3) \leq 0$

$4(6) \leq 0$

$24 \leq 0$ -------> is not ok

<u>Substitute the value of x=3 in the inequality 2</u>

$3 + 1 > 3$

$4 > 3$ --------> is ok

The value of x=3 is a solution of the compound inequality

case 5) x=8

<u>Substitute the value of x=8 in the inequality 1</u>

$4(8 + 3) \leq 0$

$4(11) \leq 0$

$44 \leq 0$ -------> is not ok

<u>Substitute the value of x=8 in the inequality 2</u>

$8 + 1 > 3$

$9 > 3$ --------> is ok

The value of x=8 is a solution of the compound inequality

<u>case 6)</u> x=10

<u>Substitute the value of x=10 in the inequality 1</u>

$4(10 + 3) \leq 0$

$4(13) \leq 0$

$52 \leq 0$ -------> is not ok

<u>Substitute the value of x=10 in the inequality 2</u>

$10+ 1 > 3$

$11 > 3$ --------> is ok

The value of x=10 is a solution of the compound inequality

therefore

<u>the answer is</u>

[-6,-3,3,8,10]

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