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3 years ago

Given circle B below, find the value of x.

Mathematics

2 answers:

marusya05 [52]3 years ago

7 0

Answer:

Step-by-step explanation:

is B correct?

blsea [12.9K]3 years ago

6 0

Answer: b is correct

Step-by-step explanation:

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How would I solve 20 I’m so confused it geometry

goldenfox [79]

Answer:

m∆IED= 83°

m∆JFG= 87°

Step by step

m∆IED is 97° so if you subtract 97 from 180(a straight line) you would get 83°

and the same for m∆JFG if you subtract 93 from 180 you get 87°

hope this helps

6 0

2 years ago

A fence post is 7 feet tall. A chain is attached to the top of the fence post and fastened to the ground 5 feet from the base of

Vanyuwa [196]

The length of the chain is 2 feet

5 0

3 years ago

Write equivalent fractions for 3^3 bottom 4 and 1^2 bottom 3

Rufina [12.5K]

$\tt \dfrac{3^3}{4}=\dfrac{27}{4}\\\\\dfrac{1^2}{3}=\dfrac{1}{3}$

5 0

2 years ago

Read 2 more answers

Last week jackson babysits 1 1/2 times as many hours as he babysat this week . last week jackson babysat 12 hours .how many hour

Stells [14]

Multiply fraction (1 1/2) by the # of hours last week (12).

= 1 1/2 * 12
change to improper fraction
= 3/2 * 12
multiply numerators
= (3*12)/2
= 36/2
= 18 hours

ANSWER: D) 18 H

Hope this helps! :)

5 0

4 years ago

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