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2 years ago

HELP IM TIMES FOR 20 MINUTES! IVE BEEN ON THIS FOR THE LONGEST!

Mathematics

1 answer:

Misha Larkins [42]2 years ago

4 0

Answer:

Point d

Step-by-step explanation:

because the x is -2 and the y is 3 so you look for the x and then the y

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PLEASE HELP WILL GIVE THANKS AND FIVE STARS IF RIGHT ANSWER

Vika [28.1K]

(x+4)^2 / 9 - (y+3)^2 / 16 = 1

a^2 = 16 and b^2 = 9

a = +4 and -4

b = +3 and -3

Center is (-4, -3)

Vertices is (-4 + a, -3) and (-4 - a, -3)

Vertices is (-1, -3) and (-7, -3)

8 0

3 years ago

sylvia has $100. She spent 4/10 of her money on a jacket and 20/100 of her money on jeans. What fraction of her money did sylvia

8_murik_8 [283]

60/100 that's what I got .

4 0

3 years ago

Read 2 more answers

The circumference of Mars is 7.00 x 107. Which planet is larger, and by how many feet? Write the answer in both standard and sci

Rasek [7]

Answer:

Step-by-step explanation:

the....

6 0

2 years ago

Any one that knows this please help thanks!

nignag [31]

You just need to plug the values: you have to multiply $x$ and $y^2$ , and then divide the result by -3.

So, if $y=2$ , we have $y^2=4$

Which means that $xy^2 = 5\cdot 4 = 20$

And the final answer is $\frac{20}{-3}$

3 0

2 years ago

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0

2 years ago