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3 years ago

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 how do you solve? thank you

zepelin [54]

Answer:

x = 2

Step-by-step explanation:

Exponential Equations

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=.................. 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