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san4es73 [151]
3 years ago
5

Help ASAP Brainliest!

Mathematics
2 answers:
leva [86]3 years ago
8 0

Answer:

(3,-3)

Step-by-step explanation:

torisob [31]3 years ago
3 0
3,-3 is the answer to your question
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Which of the following is closest to 27.8 × 9.6? A. 280 B. 300 C. 2,800 D. 3,000
Alisiya [41]
27.8 × 9.6 is equal to <span>266.88</span><span>
So, the closest would be A. 280.</span>
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balu736 [363]

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The answer for this problem is at the bottom of picture.

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What does the slope of the graph
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The slope of a graph is rise over run. So count how many units up you move and put that over how many units right you move. If you are moving down or to the left it would be negative.
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3 years ago
What is the product of (-4)(7) and why?
Evgesh-ka [11]

Answer:

The product of (-4)(7) is -28

Step-by-step explanation:

First you will multiply the number together.

This gets you -28.

The reason it is a negative number is because when you multiply a positive and a negative number together, you get a negative number.

Whereas, if it were two negative numbers, (-4)(-7), or two positive numbers,

(4)(7), the answer would have been positive, 28.

3 0
3 years ago
A car is driving away from a crosswalk. The formula d = t2 + 2t expresses the car's distance from the crosswalk in feet, d, in t
alukav5142 [94]

Answer:

8 feet per second.

Step-by-step explanation:

We have been given that a car is driving away from a crosswalk. The formula d=t^2+2t expresses the car's distance from the crosswalk in feet, d, in terms of the number of seconds, t, since the car started moving.

We will use average change formula to solve our given problem.

\text{Average change}=\frac{f(b)-f(a)}{b-a}

\text{Average change}=\frac{d(5)-d(1)}{5-1}

\text{Average change}=\frac{(5)^2+2(5)-((1)^2+2(1))}{4}

\text{Average change}=\frac{25+10-(1+2)}{4}

\text{Average change}=\frac{35-3}{4}\\\\\text{Average change}=\frac{32}{4}\\\\\text{Average change}=8

Therefore, the the car's average speed over the given interval of time would be 8 feet per second.

8 0
3 years ago
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