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3 years ago

What is four squared cubed

Mathematics

2 answers:

DiKsa [7]3 years ago

7 0

Since 4 squared is 4*4=16
16 cubed should be 16*16*16=4096

sesenic [268]3 years ago

5 0

Its exponents so u do 16x16x16= 4096

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natulia [17]

A)
Let c be the number of right guesses.
the probability of c answers =.25 and probability of the wrong answer is 1-.25=.75
c distribution is in the parameter of (20, .25).
b) mean = 20 * .25== 5
c) p of c guess =.25 ^20 =9.095* 10 ^-13
d) p of 10 c guess =.014 or 1.4% it is not possible for him to have 10 right guess.

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3 years ago

Use the given transformation to evaluate the integral. (15x + 15y) dA R , where R is the parallelogram with vertices (−1, 4), (1

MA_775_DIABLO [31]

Answer:

$\int_R 15x+15y dA = \frac{8}{16875}$

Step-by-step explanation:

Recall the following: x = 15u+15v, y = -60u+15v. So, x-y = 75u. Then u = (x-y)/75. 4x+y = 75v. Then v = (4x+y)/75.

We will see how this transformation maps the region R to a new region in the u-v domain. To do so, we will see where the transformation maps the vertices of the region.

(-1,4) -> ((-1-4)/75,(4(-1)+4)/75) = (-1/15, 0)

(1,-4)->(1/15,0)

(3,-2)->(1/15,2/15)

(1,6)->(-1/15,2/15)

That is, the new region in the u-v domain is a rectangle where $\frac{-1}{15}\leq u \leq \frac{1}{15}, 0\leq v \leq \frac{2}{15}$ .

We will calculate the jacobian of the change variables. That is

$\left |\begin{matrix} \frac{du}{dx}& \frac{du}{dy}\\ \frac{dv}{dx}& \frac{dv}{dy}\end{matrix}\right|$ (we are calculating the determinant of this matrix). The matrix is

$\left |\begin{matrix} \frac{1}{75}& \frac{-1}{75}\\ \frac{4}{75}& \frac{1}{75}\end{matrix}\right|=(\frac{1}{75^2})(1+4) = \frac{1}{15\cdot 75}$ (the in-between calculations are omitted).

We will, finally, do the calculations.

Recall that

$15x+15y = 15(15u+15v) + 15(-60u+15v) = (15^2-15\cdot 60 )u+2\cdot 15^2v = 15^2(-3)u+2\cdot 15^2 v$

We will use the change of variables theorem. So,

$\int_R 15x+15y dA = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}} 15^2(-3)u+2\cdot 15^2 v \cdot (\frac{1}{15^2\cdot 5}) dv du = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu$

This si because we are expressing the original integral in the new variables. We must multiply by the jacobian to guarantee that the change of variables doesn't affect the value of the integral. Then,

$\int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu = \int_{\frac{-1}{15}}^{\frac{1}{15}}\frac{-3}{5}u\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}}du = \frac{-3}{5}\left.\frac{u^2}{2}\right|_{\frac{-1}{15}}^{\frac{1}{15}}\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}} = \frac{8}{16875}$

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4 years ago

A costume designer bought 2 1/2 yards of black silk, 3 yards6 inches of white silk, and 6 1/3 yards of blue silk how many yards

Alexus [3.1K]

2 1/2 + 3 1/6 + 6 1/3
12 yards

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3 years ago