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2 years ago

I need help on this question can someone help.

Mathematics

1 answer:

leva [86]2 years ago

4 0

Answer:

G. (8,6)

Step-by-step explanation:

remember that EF is four times the length of DE

so the distance between E and D multiplied by 4 should equal F

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If you pay $35 for a drug and the selling price is $50 what is the mark-up of this drug

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Answer:

15$

Step-by-step explanation:

15$

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3 years ago

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bogdanovich [222]

The 3rd answer, is correct.

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3 years ago

What is the area of the triangle?

klemol [59]

A = 1/2 bh = 1/2(8)(6) = 48/2 = 24

answer
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24 square units

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3 years ago

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Julie tips the ball over the net in a volleyball game. The graph shows the quadratic function that represents the height of the

Andrew [12]

We are given a graph for the height of the ball with respect to horizontal distnace x of the ball from Julie.

From the graph, we can see the position of Julie is at (0,0). And starting height of the ball when thrown is 6 feet.

Then it moved to the maximum height of 14 feet. And hit to the ground when it has position on x-axis is 3.5 feet.

<em>Because we can see the parabola is crossing x-axis at 3.5 on the right.</em>

<em>That represents distance of the ball from Julie when it hit to the ground.</em>

<h3>Therefore, it has a horizontal distance of 3.5 feet from Julie.</h3>

6 0

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

2 years ago