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Ira Lisetskai [31]
3 years ago
7

PLEASE HELP ME.............

Mathematics
1 answer:
Brums [2.3K]3 years ago
5 0

Answer:

g(x)=3x^2

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5^(2x-1)+5^(x+1)=250<br> how do you solve?<br> thank you
zepelin [54]

Answer:

<em>x = 2</em>

Step-by-step explanation:

<u>Exponential Equations</u>

Solve:

5^{2x-1}+5^{x+1}=250

Separate each exponential:

5^{2x}5^{-1}+5^{x}5^{1}=250

Operating:

\displaystyle \frac{5^{2x}}{5}+5^{x}5=250

Multiplying by 5:

5^{2x}+25\cdot5^x=1250

Rearranging:

5^{2x}+25\cdot5^x-1250=0

Recall that:

5^{2x}=(5^{x})^2

(5^{x})^2+25\cdot5^x-1250=0

Calling

y=5^{x}:

y^2+25y-1250=0

Factoring:

(y-25)(y+50)=0

There are two possible solutions:

y=25

y=-50

Since

y=5^{x}

y cannot be negative, thus:

5^{x}=25=5^2

The solution is:

x = 2

8 0
3 years ago
You have a basketball trophy in your room. The ball on the trophy has a diameter of 9 inches. What the volume of the ball?
saveliy_v [14]

Answer:

381.51 in^3

Step-by-step explanation:

Volume of a sphere = 4/3 x pi x r^3

r = 9/2 = 4.5

4/3 x 3.14 x 4.5^3 = 381.51 in^3

7 0
2 years ago
Drugs+Kids=..................<br><br> Very tough question guys
ycow [4]

Answer:

= desaster

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po
emmasim [6.3K]
\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA
\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k
\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j

The cross product is

\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j

So, the flux is given by

\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA
\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du
\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv
\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv

where t=-\sin v in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi
5 0
3 years ago
Help me please! Posting a lot of them lol!
Vaselesa [24]

Answer:

2n -4

Step-by-step explanation:

n+8 + n-12

Combine like terms

n+n+   8-12

2n -4

3 0
3 years ago
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