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3 years ago

What is the midpoint of the segment below?

Mathematics

1 answer:

ZanzabumX [31]3 years ago

5 0

Answer:

Step-by-step explanation:

(5+(-4))/2 = 1/2 or 0.5

(-7 + 6)/2 = -1/2 or -0.5

the solution is D

(0.5, -0.5)

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−30=5(x+1) please simplify this for khan academy. org

brilliants [131]

Answer:

-7 =x

Step-by-step explanation:

−30=5(x+1)

Divide each side by 5

−30/5=5/5 (x+1)

-6 = x+1

Subtract 1 from each side

-6-1 = x+1-1

-7 =x

4 0

3 years ago

Read 2 more answers

Simplify the algebraic expression: (Combine like terms)<br><br> 25 + 3x + 16y - 15 + x - 10y

Maslowich

10+4x+6y because 25-15 3+1 and 16-10

4 0

3 years ago

Read 2 more answers

Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

4 0

3 years ago

What is the meaning of the unknown factor and quotient

konstantin123 [22]

Unknown factor is multiplication and then quotient is division

5 0

3 years ago

Match the vocabulary terms to the correct definitions.

PtichkaEL [24]

1. y axis can repeat/x cannot repeat

2. Parabola symmetrically across y-axis

3 0

3 years ago