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crimeas [40]
3 years ago
6

The wheels of a monster truck are 66 inches tall. Find the distance the monster truck travels when the tires make one 360-degree

rotation
Mathematics
1 answer:
vladimir2022 [97]3 years ago
8 0

Answer:

  About 207 inches

Step-by-step explanation:

The circumference of a circle is given by the formula ...

  C = πd . . . . . where d represents the diameter

The circumference of the tire is the distance the truck will move when the tire makes one revolution. That distance is ...

  C ≈ 3.1416×(66 in) ≈ 207.35

The truck will travel about 207 inches for each revolution of the tire.

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Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the
ludmilkaskok [199]
By Stokes' theorem,

\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S

where \mathcal C is the circular boundary of the hemisphere \mathcal M in the y-z plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle

where 0\le t\le2\pi. Then the line integral is

\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt
=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi

We can check this result by evaluating the equivalent surface integral. We have

\nabla\times\mathbf f=\langle1,0,0\rangle

and we can parameterize \mathcal M by

\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle

so that

\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv

where 0\le v\le\dfrac\pi2 and 0\le u\le2\pi. Then,

\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi

as expected.
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3 years ago
Define zero slope of line in a graph
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It would be a straight line
7 0
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Suppose you had to find 4 x 57. How are the expanded algorithm and the standard algorithm alike? And how are they different?
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The standard algorithm is given as follows:

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4 0
3 years ago
Write an expression to represent the following:
Y_Kistochka [10]
The answer would be: 5 + 9u
7 0
3 years ago
Omg, I need help! A builder is buying property where she can build new houses. The line plot shows the sizes for each house. 1/6
klemol [59]

Answer:

Average size of the lots = ⅓ acre

Step-by-step explanation:

The question incomplete without specifying what we are to determine.

Question:A builder is buying property where she can build new houses. The line plot shows the sizes for each house. 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X's. Organize the information in a line plot. What is the average size of the lots? _________ acre

Help anyone?

Solution:

We are asked to organize the information in a line plot. See attachment for the line plot.

Given: 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X'sIn no particular order, the sizes of the lots are:1/6, 1/6, 1/6, 1/6, 1/6, 1/6, 1/3, 1/3, 1/3, 1/2, 1/2, 1/2, 1/2, 1/2 and 1/2 acre.

Let's count the number of lot for each size given.

For 1/6: there are 6 X's on the line plot of 1/6 number of lot for 1/6

= the lot × number of times it occurs

= (1/6) × 6 = 6/6 = 1 acre

For 1/3: there are 3 X's on the line plot of 1/3 number of lot for 1/3 = the lot × number of times it occurs

= (1/3) × 3 = 3/3 = 1 acre

For 1/2: there are 6 X's on the line plot of 1/2 number of lot for 1/2

= the lot × number of times it occurs

= (1/2) × 6 = 6/2= 3 acres

To find average size of the lots, we would sum all lot for each given size then divide by the total number of lots given.Sum of all lot for each given size = 1+1+3

Sum of all lot for each given size = 5

The total number of lots given = 15

Average size of the lots = 5/15 = 1/3

Average size of the lots = ⅓ acre

7 0
3 years ago
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