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4 years ago

What are the co-vertices of the following graph?

Mathematics

1 answer:

Allushta [10]4 years ago

3 0

Answer:

(±1, -2)

Step-by-step explanation:

The co-vertices are the end points of the minor axis. The center is at (0, -2) and the minor axis is 2 units long, so the co-vertices are ...

(-1, -2) and (1, -2)

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Solve for x round your answer to the nearest tenth

Citrus2011 [14]

Answer:

57.9

Step-by-step explanation:

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3 years ago

Evaluate the integral by making an appropriate change of variables.

VARVARA [1.3K]

By inspecting the integrand, the "obvious" choice for substitution would be

u = y + x

v = y - x

Solving for x and y, we would have

x = (u - v)/2

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

-(u - v)/2 ≤ v ≤ (u + v)/2

-u + v ≤ 2v ≤ u + v

-u ≤ v ≤ u

So, the integral becomes

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

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

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

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

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3 years ago

How can I learn to round better in math

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3 years ago

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Keith_Richards [23]

Answer:b=2.5555555555...

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3 years ago

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Alexandra [31]

Answer:

Statement 4 in true.

Step-by-step explanation:

1. f(-2)=(-3*(-2)+2)^2+(-2)=(6+2)^2+(-2)=8^2-2=62, FALSE

2. f(-5)=(-3*(-5)+2)^2+(-5)=(15+2)^2+(-5)=17^2+(-5)=289-5=284, FALSE

3. f(3)=(-3*3+2)^2+3=(-9+2)^2+3=(-7)^2+3=49+3=52, FALSE

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3 years ago

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