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Romashka-Z-Leto [24]
3 years ago
10

Find the area of a rectangle with a width of x^3 +5x-9 and a length of x^2 +8​

Mathematics
1 answer:
Effectus [21]3 years ago
5 0

Awnser: The length is 5 x 3 = 15 inches.

Explination: Multiplied by the width of 3 inches, yields 45 in2.

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My 4 time posting this can anyone please help me please 14 and 16
Katena32 [7]

Answer:

14. 780

16. 274

hopes this helps have a great night

Step-by-step explanation:

5 0
3 years ago
I need help<br> (percents)
Reptile [31]
CHOICE 1 $1250+5.5% of sales
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CHOICE 2  $2600=2% of sales
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So even though more is earned through percentage of sales in CHOICE 1, it doesn't make up for the larger monthly salary of CHOICE 2
8 0
3 years ago
Nonpermissable replacement for a in 4/-9a
Temka [501]
If a=0, then the denominator is equal to 0.  Since you cannot divide by zero, you are not allowed to do this.
5 0
4 years ago
Use stokes' theorem to evaluate c f · dr where c is oriented counterclockwise as viewed from above. f(x, y, z = xyi + 5zj + 7yk,
Helga [31]
The intersection can be parameterized by

C:=\mathbf r(t)=\begin{cases}x(t)=6\cos t\\y(t)=6\sin t\\z(t)=5-6\cos t\end{cases}

with 0\le t.

By Stoke's theorem, the integral of \mathbf f(x,y,z)=xy\,\mathbf i+5z\,\mathbf j+7y\,\mathbf k along C is equivalent to

\displaystyle\int_C\mathbf f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r(t)=\iint_S\nabla\times\mathbf f\,\mathrm dS

where S is the region bounded by C. The line integral reduces to

\displaystyle\int_0^{2\pi}(36\sin t\cos t\,\mathbf i+(25-30\cos t)\,\mathbf j+42\sin t\,\mathbf k)\cdot(-6\sin t\,\mathbf i+6\cos t\,\mathbf j+6\sin t\,\mathbf k)\,\mathrm dt
=\displaystyle\int_0^{2\pi}(54(\cos3t-\cos t)-30(3\cos2t-5\cos t+3)+(126-126\cos2t)\,\mathrm dt
=\displaystyle\int_0^{2\pi}(36+96\cos t-216\cos2t+54\cos3t)\,\mathrm dt
=72\pi
4 0
4 years ago
6. What number is not a prime number or a perfect square that lies between
Serjik [45]
The number would be 22
7 0
3 years ago
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