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Firlakuza [10]
3 years ago
6

Dima called her friend to tell her that she saved 30% on her new skirt at a discount store. Her friend told her that she could h

ave gotten a better deal at a different store that was advertising a sale of off all clothing. If the original price of Dima’s skirt was $54, how much more could she have saved at the store her friend suggested?
$1.11
$1.80
$36.00
$37.80
Mathematics
2 answers:
Mkey [24]3 years ago
8 0

Answer:

The correct answer is $1.80

Ket [755]3 years ago
4 0

Answer:

$1.80

Step-by-step explanation:

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Describe what the graph of the sounds at a rock concert look like.
Romashka-Z-Leto [24]

The graph would be similar to a Sin or Cos graph with a vertical shift. As the music gets louder the graph will go up and as the music and crowd become quieter, the graph will go down. Going up and down until the end of the concert where it will be 0.


Also side note, are you by chance in Trig 01 Summer course? If so, I am in the same class haha.

3 0
3 years ago
Evaluate the expression for y=-14
Daniel [21]
Y=0-14 i think is the amswer
5 0
4 years ago
One tenths times one decimal point six
barxatty [35]
1/10 x 1.6

First make 1/10 over 100
1/10(10/10) = 10/100, which = 0.10

0.1 x 1.6 = 0.16

0.16, or 16/100, is your answer

hope this helps
6 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
8. Write the equation of a line that is perpendicular to the given line and that passes through the
avanturin [10]
Y - 3 = 8/3(x + 2)...the slope here is 8/3. A perpendicular line will have a negative reciprocal slope. All that means is " flip " the slope and change the sign. So our perpendicular line will have a slope of -3/8

y - y1 = m(x - x1)
slope(m) = -3/8
(-2,3)....x1 = -2 and y1 = 3
now we sub
y - 3 = - 3/8(x - (-2)...not done yet
y - 3 = -3/8(x + 2) <===

8 0
3 years ago
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