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Ulleksa [173]
3 years ago
8

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
5 0
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:

algebraic expression for 180 degree clockwise rotation about the origin (x,y) → (-y, x) algebraic expression for 270 degree clockwise rotation about the origin equals a 270 degree counterclockwise rotation 90 degree clockwise rotation equals a 90 degree counterclockwise rotation

Step-by-step explanation:

hope this helps

have an awesome day -TJ

3 0
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:

5 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
3 years ago
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