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2 years ago

11

Help me on this please

1 answer:

gulaghasi [49]2 years ago

5 0

I think +5
Not completely sure

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Coach Rivas can spend up to $750 on 30 swimsuits for the swim team. The inequality shown can be used to find the maximum amount

777dan777 [17]

Answer:

$0 < p ≤ $25

Step-by-step explanation:

We know that coach Rivas can spend up to $750 on 30 swimsuits.

This means that the maximum cost that the coach can afford to pay is $750, then if the cost for the 30 swimsuits is C, we have the inequality:

C ≤ $750

Now, if each swimsuit costs p, then 30 of them costs 30 times p, then the cost of the swimsuits is:

C = 30*p

Then we have the inequality:

30*p ≤ $750.

To find the possible values of p, we just need to isolate p in one side of the inequality.

So we can divide both sides by 30 to get:

(30*p)/30 ≤ $750/30

p ≤ $25

And we also should add the restriction:

$0 < p ≤ $25

Because a swimsuit can not cost 0 dollars or less than that.

Then the inequality that represents the possible values of p is:

$0 < p ≤ $25

6 0

3 years ago

Help please. step by step explanation appreciated

olya-2409 [2.1K]

Answer:

Step-by-step explanation:

Are the options just A and B? If so , the answer is B. the volume is the width, length, times height, so 8 times 1 times 3 is 24, and that would fit the answer.

5 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP

butalik [34]

Answer: Dilation followed by a rotation & they are similar.

Step-by-step explanation:

3 0

2 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

2 years ago

In this equation What is 77+23

kondaur [170]

The answer is 100 I’m sure

3 0

3 years ago

Read 2 more answers