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Nataly [62]
4 years ago
9

Last week, a skate shop originally sold a skateboard for $120. Today, there is a sale and the same skateboard costs $96. What is

the percentage of the discount?
Mathematics
1 answer:
Alex Ar [27]4 years ago
8 0
80%. That is the answer
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PLEASE HELP! The diagonals of a trapezoid are equal.<br><br> always<br> sometimes<br> never
nikdorinn [45]

The diagonals of a trapezoid are equal sometimes,

and they're unequal sometimes.


The diagonals are equal only in an "Isosceles trapezoid" . . .

where the non-parallel sides have equal lengths.

4 0
3 years ago
A building is 200ft long and 140 ft wide a 1/500 model is built of the building how long and how wide is the model?
Katena32 [7]
Let's start with the length. All you need to do is multiply is by 1/500.
You could also find the decimal equivalent to 1/500.
1/500= .002
200×.002= .4
So length of the model is .4 ft.
The width:
.002×140=.28
The width of the model is .28 ft.
The length of the model is .4 ft and the width of the model is .28 ft.

4 0
3 years ago
The width of a rectangle is y feet long. The rectangle is 4 feet longer than it is wide. What is the area of the rectangle?
mario62 [17]

Answer:

if breadth is y feet long

let length be y+4

Step-by-step explanation:

therefore,

area=(4+y)*y

area=4y+y2

4 0
4 years ago
Read 2 more answers
Help Me With These Please
anzhelika [568]
10×sine of 60 degrees=8.66 which is the same as B. you can check by finding the square root of 3 and then multiplying that by 5.
6 0
3 years ago
Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
saw5 [17]
Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

with \mathbf f=(y+7\sin x,z^2+9\cos y,x^3).

By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


\nabla\times\mathbf f=(-2z,-3x^2,-1)

so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv


where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi<span />
4 0
3 years ago
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