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3 years ago

How do u find the point in a graph if they are not given to u when u solving changing y over changing x

1 answer:

3 0

U would replace x with 0 in the equation and solve and then do the same but with y and those are ur x and y values of ur point
Ex: 2x+5y=15
2(0) +5y=15
5y=15
Y=3
2x+5(0)=15
2x=15
X=7.5
So ur point would be (7.5,3)

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omeli [17]

it would be C i think hope this helps

5 0

3 years ago

Read 2 more answers

PLEASE HELP 30 POINTS

mote1985 [20]

Answer: Maggie is 29 years old

Step-by-step explanation:

12*3=36

36-7=29

29 + 12 = 41

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3 years ago

BRAINLIEST!! LOTS of POINTS

zubka84 [21]

Answer:

-0.2

Step-by-step explanation:

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3 years ago

In a game, you pilot a spaceship and gain a point for every asteroid you

alina1380 [7]

Answer:

y = p+94

Step-by-step explanation:

If you already have 94 points you can't get lower than that.

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3 years ago

Use spherical coordinates to find the volume of the region that lies outside the cone z = p x 2 + y 2 but inside the sphere x 2

Harman [31]

I assume the cone has equation $z=\sqrt{x^2+y^2}$ (i.e. the upper half of the infinite cone given by $z^2=x^2+y^2$ ). Take

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

The volume of the described region (call it $R$ ) is

$\displaystyle\iiint_R\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\sqrt2}\int_{\pi/4}^\pi\rho^2\sin\varphi\,\mathrm d\varphi\,\mathrm d\rho\,\mathrm d\theta$

The limits on $\theta$ and $\rho$ should be obvious. The lower limit on $\varphi$ is obtained by first determining the intersection of the cone and sphere lies in the cylinder $x^2+y^2=1$ . The distance between the central axis of the cone and this intersection is 1. The sphere has radius $\sqrt2$ . Then $\varphi$ satisfies

$\sin\varphi=\dfrac1{\sqrt2}\implies\varphi=\dfrac\pi4$

(I've added a picture to better demonstrate this)

Computing the integral is trivial. We have

$\displaystyle2\pi\left(\int_0^{\sqrt2}\rho^2\,\mathrm d\rho\right)\left(\int_{\pi/4}^\pi\sin\varphi\,\mathrm d\varphi\right)=\boxed{\frac43(1+\sqrt2)\pi}$

4 0

3 years ago