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yKpoI14uk [10]
2 years ago
8

Will give brainleiest to best answer of 2 please help!!

Mathematics
2 answers:
LenaWriter [7]2 years ago
6 0
The second one I think is 6
Paha777 [63]2 years ago
5 0

A.

For Deshaun: $75 - $10x

For Ann: $90 - $15x

B.

$75 - $10x = $90 - $15x

<em>I believe this is right, check over to make sure. I may be wrong, and sorry if I am.</em>

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What is it the following a perfect square A49 B121 C75 D64​
guajiro [1.7K]

Answer:

D is a perfect square

Step-by-step explanation:

a perfect square is a squared number that has a square root of an even number like sqrt(64) = 8 and 8 is an even number making it a perfect square

7 0
2 years ago
Your neighbor burns 100 Calories in 10 minutes​ cross-country skiing. There is a proportional relationship between Calories burn
Zinaida [17]

Answer:

10

Step-by-step explanation:

For every one minute 10 calories are being burned

4 0
3 years ago
Can anyone help me x55x5x5x56x6x7x7= ?<br><br> this is for my gf
77julia77 [94]

Answer:x 55x5x5x56x6x7x74 4,527,600

3 0
3 years ago
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umka2103 [35]
32 floors
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2 years ago
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Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos &#10;y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
3 years ago
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