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lesantik [10]
3 years ago
7

Jaheem has a goal running a total of 125 miles this month. Each day that he ran, he ran 7 miles. Which expression could Jaheem u

se to determine how many miles he has left to run after running for d days?
x a.
125 – 7d


x b.
7d + 125


x c. fraction numerator 125 over denominator 7 d end fraction

x d.
7d
Mathematics
1 answer:
VashaNatasha [74]3 years ago
5 0

Answer:

a. 125 – 7d

Step-by-step explanation:

Let d be the number of days.

<u>Given the following data;</u>

Total distance (month) = 125 miles

Distance covered per day = 7 miles

To determine how many miles he has left to run after running for "d" days, this mathematical expression can be used;

Total distance - distance covered per day = number of miles remaining

125 - 7d

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Answer:

  9t^3 +t^2

Step-by-step explanation:

The perimeter of the figure is the sum of the lengths of the sides. The side lengths are represented by the polynomials shown, so the perimeter (P) is their sum:

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Rearranging to group like terms:

  P = (4t^3 +4t^3 +t^3) + (t^2 -t^2 +t^2) + (-5 -5 +9 -11 +12)

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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,

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Complete the square in the denominator:

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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
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