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Sever21 [200]
3 years ago
7

The volume V of a rectangular prism with length l, width w, and height h is v equals l times w times h. What is the width of a r

ectangular prism with volume 450 cubic inches, length 20 inches, and height 7.5 inches?
Mathematics
2 answers:
erastovalidia [21]3 years ago
8 0
450=w•20•7.5
^
450= w150
——- ———
150. 150

3=w
Iteru [2.4K]3 years ago
6 0

Answer:

The answer is 3 inch

Step-by-step explanation:

To start off, the equation you want to use the formula V=whl

Next, you simply plug in the values you already know, so you have 450=w*20*7.5

After you do that you multiply 20 times 7.5 to get which is 150, which leaves you with 450= w *150 to solve for w, you need to isolate it. To do that you divide 450 by 150. You are left with the answer of 3.

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Find the moment of inertia about the y-axis of the thin semicirular region of constant density
arlik [135]

The moment of inertia about the y-axis of the thin semicircular region of constant density is given below.

\rm I_y = \dfrac{1}{8} \times \pi r^4

<h3>What is rotational inertia?</h3>

Any item that can be turned has rotational inertia as a quality. It's a scalar value that indicates how complex it is to adjust an object's rotational velocity around a certain axis.

Then the moment of inertia about the y-axis of the thin semicircular region of constant density will be

\rm I_x = \int y^2 dA\\\\I_y = \int x^2 dA

x = r cos θ

y = r sin θ

dA = r dr dθ

Then the moment of inertia about the x-axis will be

\rm I_x = \int _0^r \int _0^{\pi}  (r\sin \theta )^2  \ r \  dr \  d\theta\\\\\rm I_x = \int _0^r \int _0^{\pi}  r^3 \sin ^2\theta  \  dr \  d\theta

On integration, we have

\rm I_x = \dfrac{1}{8} \times \pi r^4

Then the moment of inertia about the y-axis will be

\rm I_y = \int _0^r \int _0^{\pi}  (r\cos\theta )^2  \ r \  dr \  d\theta\\\\\rm I_y = \int _0^r \int _0^{\pi}  r^3 \cos ^2\theta  \  dr \  d\theta

On integration, we have

\rm I_y = \dfrac{1}{8} \times \pi r^4

Then the moment of inertia about O will be

\rm I_o = I_x + I_y\\\\I_o = \dfrac{1}{8} \times \pi r^4 + \dfrac{1}{8} \times \pi r^4\\\\I_o = \dfrac{1}{4} \times \pi r^4

More about the rotational inertia link is given below.

brainly.com/question/22513079

#SPJ4

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2 years ago
You are given odds of 7 to 8 in favor of winning a bet, what is the probability of winning the bet?
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The probability of winning the bet is very likely.

Hope this helps!

3 0
4 years ago
Read 2 more answers
Apply Gaussian quadrature with n = 4 to approximate integrate sin x^2dx from 1 to 5
maks197457 [2]
First, recall that Gaussian quadrature is based around integrating a function over the interval [-1,1], so transform the function argument accordingly to change the integral over [1,5] to an equivalent one over [-1,1].

x=2t+3\iff t=\dfrac x2-\dfrac32\implies2\mathrm dt=\mathrm dx
x=1\implies t=\dfrac{2-6}4=-1
x=5\implies t=\dfrac{10-6}4=1

So,

\displaystyle\int_{x=1}^{x=5}\sin x^2\,\mathrm dx=\displaystyle2\int_{t=-1}^{t=1}\sin(2t+3)^2\,\mathrm dt

Let f(t)=2\sin(2t+3)^2. With n=4, we're looking for coefficients c_i and nodes x_i, with 1\le i\le4, such that

\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx c_1f(x_1)+\cdots+c_4f(x_4)

You can either try solving for each with the help of a calculator, or look up the values of the weights and nodes (they're extensively tabulated, and I'll include a link to one such reference).

Using the quadrature, we then have

\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.3749f(-0.8611)+0.6521f(-0.3400)+0.6521f(0.3400)+0.3749f-0.8611)
\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.5790
4 0
3 years ago
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