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3 years ago

I need the answer and work

Mathematics

1 answer:

ad-work [718]3 years ago

7 0

A) 2x +3

Plug in the x

2(4.1) + 3

8.2+ 3= 11.2

A is 11.2

B) 16-5y

Plug in the y

16 - 5(2.3)

16- 11.5= 4.5

B is 4.5

C) x+y

Plug in x and y

4.1 + 2.3= 6.4

C is 6.4

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I am confused with everything

notsponge [240]

I think it's x^3+2x^2-4

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3 years ago

350 divided by 10 to the 3rd power

Sergio [31]

Answer:

<h2>0.35</h2><h2>Step-by-step explanation:</h2>

350 divided by 10 to the 3rd power

350 : 10³ =

350 : 1000 =

0.35

7 0

3 years ago

Which is the algebraic representative for a rotation 180 degrees clockwise?

neonofarm [45]

Answer:

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Step-by-step explanation:

hope this helps

have an awesome day -TJ

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2 years ago

I don’t know if it negative or zero please help me

Paraphin [41]

Answer:

i think it is negative

Step-by-step explanation:

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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