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galben [10]
2 years ago
14

A handyman buys lumber to fix a fence. The lumber is $562 before his 20% discount. There is 6% sales tax on his purchase, but hi

s credit card gives him an additional 2% discount. How much does he spend in all?
Mathematics
1 answer:
kari74 [83]2 years ago
4 0
Should be 445.10. I just took the percentages of all out of 562 and added and subtracted as necessary.
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What percent of a meter is 30 cm
Alexxx [7]
A meter = 10000 cm

30cm/1m
30/10000
=0.003
x100 (for the percent)

=0.3%

6 0
3 years ago
Ross calculated the missing side length of one of these triangles using the Pythagorean Theorem. Which triangle was it?
sveta [45]
It would be triangle E
3 0
3 years ago
Read 2 more answers
Compare the ratios in Table 1 and Table 2.
Grace [21]

Answer:

3:5 is less tan 7:10

The ratio 14:20 is greater than the ratio 9:15.

The ratios in Table 1 are less than the ratios in Table 2.

Step-by-step explanation:

3:5 as a fraction is 6/10

7:10 as a fraction is 7/10

3:5 is less

3 0
3 years ago
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Find the scalar and vector projections of b onto<br> a.a = i + j + k, b = i - j + k
yarga [219]
Try this option.
1 is for scalar (result is number), 2 is for vector (result is vector).

5 0
3 years ago
Find the area of the surface. the part of the plane 4x + 4y + z = 16 that lies inside the cylinder x2 + y2 = 9
MAVERICK [17]
As a first step, convert to cylindrical coordinates. The equation x^2+y^2=9=3^2 is a clue to set

x=3r\cos\theta

y=3r\sin\theta

then have 0\le r\le1 and 0\le\theta\le2\pi. Meanwhile, the plane equation tells you to use

4x+4y+z=16\implies z=16-12r\cos\theta-12r\sin\theta

So we parameterize the part of the plane within the cylinder with the vector-valued function

\mathbf s(r,\theta)=(3r\cos\theta,3r\sin\theta,15-12r\cos\theta-12r\sin\theta)

The surface integral giving the area of the surface is


\displaystyle\iint_{\mathcal S}\mathrm dS

where \mathcal S denotes the surface in question, and the surface element \mathrm dS is

\mathrm dS=\left\|\dfrac{\partial\mathbf s}{\partial r}\times\dfrac{\partial\mathbf s}{\partial\theta}\right\|\,\mathrm dr\,\mathrm d\theta

We have

\dfrac{\partial\mathbf s}{\partial r}=(3\cos\theta,3\sin\theta,-12(\cos\theta+\sin\theta))

\dfrac{\partial\mathbf s}{\partial\theta}=(-3r\sin\theta,3r\cos\theta,12r(\sin\theta-\cos\theta))

\implies\mathrm dS=\|(36r,36r,9r\|\,\mathrm dr\,\mathrm d\theta=9\sqrt{33}r\,\mathrm dr\,\mathrm d\theta

So the area is

\displaystyle\iint_{\mathcal S}\mathrm dS=9\sqrt{33}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta=18\sqrt{33}\,\pi
4 0
3 years ago
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