Answer question! I really need help

Mathematics

1 answer:

emmasim [6.3K]3 years ago

6 0

I can't answer it the problem

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R’(-1, 9) is the image of R after a reflection in the x-axis. What are the coordinates of R?

xz_007 [3.2K]

Answer:

Step-by-step explanation:

When we reflect over x-axis then the x-coordinate doesn't change and the y-coordinate changes its sign

in means if R = (x, y) then R' = (x, -y)

so we have:

x = -1 and -y = 9

y = -9

which gives R = (-1, -9)

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3 years ago

Need HELP!!! For all, I’m truly confused

irina1246 [14]

Step-by-step explanation:

You need to know some geometry rules.

For example:

Angle sum of a triangle is equal to 180 degrees.

Straight line is equal to 180 degrees.

Other rules I have attached.

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3 years ago

Choose the best coordinate system to find the volume of the portion of the solid sphere rho <_4 that lies between the cones φ

MrRissso [65]

Answer:

So, the volume is:

$\boxed{V=\frac{128\sqrt{2}\pi}{3}}$

Step-by-step explanation:

We get the limits of integration:

$R=\left\lbrace(\rho, \varphi, \theta):\, 0\leq \rho \leq 4,\, \frac{\pi}{4}\leq \varphi\leq \frac{3\pi}{4},\, 0\leq \theta \leq 2\pi\right\rbrace$

We use the spherical coordinates and we calculate a triple integral:

$V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_0^4 \rho^2 \sin \varphi \, d\rho\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \left[\frac{\rho^3}{3}\right]_0^4\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \cdot \frac{64}{3} \, d\varphi\, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} [-\cos \varphi]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\$