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Leona [35]
3 years ago
15

5^(11x+23)=125^(7x) how to solve for x

Mathematics
1 answer:
ruslelena [56]3 years ago
8 0
Answer: x= 23/10 or x= 2.3
First... 5^11x-23=5^21x
Cancel out the 5
Secondly... 11x+23 =21x
Switch 21 and 23 to make...
11x=21x-23. Switch again to make...
11x-21x=23
Now subtract 11 and 21
-10x=-23
Divide to get
x=23/10 or x=2.3
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You need to add a -1

We can find this by setting up the equation like so

x + (-6) + 12 + 15 / 4 = 5

x + 21 = 20

x = 20 -21

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3 years ago
A triangle has side lengths of 1512, 223, and 314. What is the perimeter of the triangle? Remember that the perimeter of a trian
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Answer:

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3 years ago
Quadrilateral ABCD ​ is inscribed in a circle.
Lerok [7]
In general, the sum of the measures of the interior angles of a quadrilateral is 360. This is true for every quadrilateral. This does not help here, because there are two angles (angles B and D) we know nothing about. We only know about opposite angles A and C.

In this case, you can use another theorem.
Opposite angles of an inscribed quadrilateral are supplementary.

m<A + m<C = 180

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6 0
3 years ago
A sphere of radius r is cut by a plane h units above the equator, where
Anika [276]
Consider the top half of a sphere centered at the origin with radius r, which can be described by the equation

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and consider a plane

z=h

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\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx

Converting to polar coordinates will help make this computation easier. Set

\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

Now, the volume can be computed with the integral

\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta

You should get

\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)
5 0
3 years ago
Find the slope (4, -1) and (-2, 3)
Yuliya22 [10]
Answer: -2/3

3-(-1) over
-2-4
This equals 4/-6 which reduces to -2/3
5 0
2 years ago
Read 2 more answers
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