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vladimir2022 [97]
3 years ago
13

Determine if each function is linear or nonlinear.

Mathematics
1 answer:
Akimi4 [234]3 years ago
3 0

Answer:

Linear: y = x, y = x/2 - 3, 3x + 2 = 12

Nonlinear: y = 6/x - 2, y = 3x^3 + 5

Step-by-step explanation:

Linear functions form straight lines while nonlinear functions do not.

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Evaluate the integral by making an appropriate change of variables.
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By inspecting the integrand, the "obvious" choice for substitution would be

<em>u</em> = <em>y</em> + <em>x</em>

<em>v</em> = <em>y</em> - <em>x</em>

<em />

Solving for <em>x</em> and <em>y</em>, we would have

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<em>y</em> = (<em>u</em> + <em>v</em>)/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 <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.

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

-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>

-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2

-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>

-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>

<em />

So, the integral becomes

\displaystyle\iint_R5\cos\left(7\frac{y-x}{y+x}\right)\,\mathrm dA=\int_8^9\int_{-u}^u\frac52\cos\left(\frac{7v}u\right)\,\mathrm dv\,\mathrm du

=\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}

4 0
2 years ago
A recipe calls for the following spices: 18 teaspoon of turmeric, 18 teaspoon of ginger, and 14 teaspoon of cumin. What is the t
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Answer:

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when it's asking for a total you add the numbers together, and whatever number you get is your answer :)

sorry if it's wronggg :( !!

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