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horrorfan [7]
3 years ago
14

a pyramid has a volume of 1944 in and a rectangular base with the dimensions of 18 and 9 in what is the height of the pyramid​

Mathematics
1 answer:
Katarina [22]3 years ago
7 0

Answer:

The height is 36 in

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, a rectangular based pyramid.

Step one

Given data

Volume v =  1944 in

length of base l =  18 in

width of base w = 9 in

Height h=   ?

Step two

we know that the expression for the volume of a pyramid is given as

volume=\frac{ lwh}{3}

substituting our given data we have

volume = \frac{18*9*h}{3} \\volume = \frac{162h}{3} \\volume= 54h

but volume = 1944 in therefore

1944= 54h\\h= \frac{1944}{54} \\h= 36 in

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Let R be the region in the first quadrant of the​ xy-plane bounded by the hyperbolas xyequals​1, xyequals9​, and the lines yequa
Tema [17]

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The area can be written as

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274

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Step-by-step explanation:

x = u/v

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Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.

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  • x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4

Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).

Lets compute the partial derivates of x and y over u and v

x_u = 1/v

x_v = u*ln(v)

y_u = v

y_v = u

Therefore, the Jacobian matrix is

\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right]

and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))

In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:

\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}

Therefore,

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274

4 0
3 years ago
Can anyone help me with number 36? I’m trying to figure out how to do this am I’m having a lot of problems. The answer is in the
zubka84 [21]

You have 9i in the denominator. The goal is to have a real denominator. 9i is imaginary. i * i = -1 which is real. If you multiply the denominator by i, the denominator will be real. Since this is a fraction, we multiply the numerator and denominator by i.

\dfrac{-6 - 10i} {9i } \times \dfrac{i}{i} =

= \dfrac{(-6 - 10i)i}{(9i)i}

= \dfrac{-6i - 10i^2}{9i^2}

= \dfrac{-6i - 10(-1)}{9(-1)}

= \dfrac{-6i + 10}{-9}

= \dfrac{6i - 10}{9}

= \dfrac{-10 + 6i}{9}

= \dfrac{-10}{9} + \dfrac{6i}{9}

= \dfrac{-10}{9} + \dfrac{2i}{3}

8 0
3 years ago
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