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igor_vitrenko [27]
2 years ago
7

Helppppp plzzzzzz im almost of time i will give brainiest

Mathematics
1 answer:
Anettt [7]2 years ago
5 0
All your doing is multiplying area, so just foil each term. The answer you will get is 24w^3-30w^2+17w-6
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Help me if you understand thanks sm
kumpel [21]
First, find the slope of the line using the formula for slope:

m = (y2 - y1) / (x2 - x1)

m = (4 - (-2)) / (-3 - 2) = -6/5

Then, use the point-slope formula using a point that you were given and the slope that you calculated.

y - y1 = m(x - x1)

y - 4 = (-6/5)(x + 3)

Then, simply get y alone.

y = (-6/5)x + 2/5
3 0
2 years ago
Helpp me pls I'm struggling
Ierofanga [76]

Answer:

B. 2nd Play and B. 2nd Play

Step-by-step explanation:

#1) Going negative means going backwards.

-8<10.5

#2) Absolute value means making all numbers INSIDE the two lines positive.

I-8I = 8

I10.5I = 10.5

3 0
2 years ago
Read 2 more answers
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
At which step did Fatu make an error, if at all?
Elza [17]

Answer:

Step 3

Step-by-step explanation:

7 0
3 years ago
Help a partner out thank youuu
Katen [24]
You should just use so.crative or photo.math
6 0
3 years ago
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