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Advocard [28]
3 years ago
12

Ms. Keisha has just purchased supplies for her classroom. She bought 67 markers, 32 rulers, and one projector. The markets cost

$0.90 each, the rulers cost $1.95 each, and the projector cost $98.95. What was the total amount that she spent ?
Mathematics
2 answers:
Vladimir79 [104]3 years ago
7 0

Answer:

221.65

Step-by-step explanation:

67 x .9 = 60.3

32 = 1.95 = 62.4

so you got 60.3 + 62.4 + 98.95 = 221.65

hope this helps please make brainliest.

timofeeve [1]3 years ago
4 0

Answer: <u>$189.08</u>


Step-by-step explanation:

67 / .90= 74.44

32 / 1.95= 16.41

Projector = 98.95

74.44 + 16.41 + 98.95= <u>189.08 </u>


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Lesechka [4]

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Answer:

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There are no values of x for which this expression is undefined. The domain is all real numbers.

The minimum value of this expression is 0. There is no maximum value. The range is all numbers greater than or equal to zero.

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3 years ago
How do you rename a mixed number as a fraction khan academy.
elena-s [515]

Answer:

There's no answer choices but I can give you an example.  

Step-by-step explanation:

  1. Let's say you have \frac{25}{4}.  
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5 0
2 years ago
What’s the answer? I need help I don’t like similar triangles
Sedbober [7]
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8 0
3 years ago
Read 2 more answers
The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo
oksano4ka [1.4K]

Answer:

Required center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

Step-by-step explanation:

Given semcircles are,

y=\sqrt{1-x^2}, y=\sqrt{16-x^2} whose radious are 1 and 4 respectively.

To find center of mass, (\bar{x},\bar{y}), let density at any point is \rho and distance from the origin is r be such that,

\rho=\frac{k}{r} where k is a constant.

Mass of the lamina=m=\int\int_{D}\rho dA where A is the total region and D is curves.

then,

m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k

  • Now, x-coordinate of center of mass is \bar{y}=\frac{M_x}{m}. in polar coordinate y=r\sin\theta

\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta

=3k\int_{0}^{\pi}\sin\theta d\theta

=3k\big[-\cos\theta\big]_{0}^{\pi}

=3k\big[-\cos\pi+\cos 0\big]

=6k

Then, \bar{y}=\frac{M_x}{m}=\frac{2}{\pi}

  • y-coordinate of center of mass is \bar{x}=\frac{M_y}{m}. in polar coordinate x=r\cos\theta

\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta

=3k\int_{0}^{\pi}\cos\theta d\theta

=3k\big[\sin\theta\big]_{0}^{\pi}

=3k\big[\sin\pi-\sin 0\big]

=0

Then, \bar{x}=\frac{M_y}{m}=0

Hence center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

3 0
3 years ago
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Oxana [17]

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Step-by-step explanation:

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Replacing with the values given and calculated:

V=1/3 π (3.09)^2 (17.5)

V = 174.71 cubic meters

4 0
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