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3 years ago

Please help solve both due soon

Mathematics

1 answer:

Mariana [72]3 years ago

7 0

Answer:

x is 7

EF is 10

FG is 12

Step-by-step explanation:

$EF + FG = EG \\ (4x - 18) + (3x - 9) = 22 \\ 7x - 27 = 22 \\ 7x = 49 \\ x = 7$

then substitute:

$EF = 4x - 18 \\ EF = 4(7) - 18 \\EF = 28 - 18 \\ EF = 10$

$FG = 3x - 9 \\ FG = 3(7) - 9 \\FG = 21 - 9 \\ FG = 12$

First question:

BC + BA = CA

BC = CA - BA

BC = 36 - 9

BC = 27

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Evaluate the integral by making an appropriate change of variables.

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u = y + x

v = y - x

Solving for x and y, we would have

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y = (u + v)/2

in which case the Jacobian and its determinant are

$J=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}=\dfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}\implies|\det J|=\left|\dfrac12\right|=\dfrac12$

The trapezoid R has two of its edges on the lines x + y = 8 and x + y = 9, so right away, we have 8 ≤ u ≤ 9.

Then for v, we observe that when x = 0 (the lowest edge of R), v = y ; similarly, when y = 0 (the leftmost edge of R), v = -x. So

-x ≤ v ≤ y

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-u + v ≤ 2v ≤ u + v

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$=\displaystyle\frac52\int_8^9\frac u7(\sin7-\sin(-7))\,\mathrm du$

$=\displaystyle\frac57\sin7\int_8^9u\,\mathrm du$

$=\displaystyle\frac5{14}\sin7(9^2-8^2)=\boxed{\frac{85}{14}\sin7}$

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