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nadezda [96]
3 years ago
12

Given circle B below, find the value of x.

Mathematics
2 answers:
marusya05 [52]3 years ago
7 0

Answer:

b

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
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