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3 years ago

How do you translate the shape for 11 and 12?

Mathematics

1 answer:

Ahat [919]3 years ago

8 0

Count the points to the mirror line then continue onto the other side marking a dot on the opposite side for each corner of the shape. Then join up the dots.
Hope that helps!

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Which statement is true

AlladinOne [14]

3) an $8 delivery fee and $1.50 per litre of water

This is because the $8 is a constant baseline, then adding $1.50 times the amount of litres purchased.

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2 years ago

Given that x has a Poisson of u=5 what is the probability that x =2

mr_godi [17]

probability//

Step-by-step explanation:

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1 year ago

7/7, solve for k, explain too please

ludmilkaskok [199]

Answer:

k = 7.3333333 ...

You can round accordingly of write the fraction form

Step-by-step explanation:

4/k = 6/11

6 ÷ 4 = 1.5

11 ÷ 1.5 = 7.3333333 ... (repeating)

4/7.333 = 6/11

7 0

2 years ago

Determine the domain and the range of the relation. *Type the domain and range in number form from least to greatest

slavikrds [6]

Answer:

i dont understand what your asking

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

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