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timurjin [86]
3 years ago
5

How many quarters are there in $2? A. 2 B. 4 C. 6 D. 8

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
7 0
The answer is D there are 8 quarters in $2
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Which vectors represent the reflection of the vector <3, -7> across the x-axis? A. [3/7} B. [-3/7] C. <-3, 7> D. &lt
Alexandra [31]

Answer:

D)    < 3, 7)>

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Given that the vector < 3 , -7 >

Given the vector reflection across the x-axis

                   (x,y) → (x , -y)

The vector < 3,-7> →< 3, -(-7)>

                  < 3,-7> →< 3, 7)>

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Aleksandr-060686 [28]

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ITS C THE ANSWER IS C

Step-by-step explanation:

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The watch at Macys is $400. The store is having a sale and everything is 30% off. What is the discount amount?
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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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