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soldi70 [24.7K]
3 years ago
13

Solve for x: −1 < x + 3 < 5

Mathematics
2 answers:
Art [367]3 years ago
8 0

Answer:

B. −4 < x < 2

Step-by-step explanation:

−1 < x + 3 < 5

To make x the subject, subtract 3 across the inequality

−1 - 3 < x + 3 - 3 < 5 - 3

-4 < x < 2

sineoko [7]3 years ago
6 0

Answer:

-4 < x < 2

Step-by-step explanation:

−1 < x + 3 < 5

- 3 - 3 - 3

_____________

−4 < x < 2

I am joyous to assist you anytime.

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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
write an equation for the line with a slope of 3/2 that passes through (4,-2) (I need the answer asap)
Ierofanga [76]

The equation of the line that has a slope m is

y=mx+b.

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y=\dfrac{3}{2}x+b.

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3 years ago
In 2000, there were 7689 weekly newspapers in the United States. This s about 94% of the number of weekly newspapers in the Unit
Elina [12.6K]
In 2000, 7689 weekly papers = 94% of number of weekly papers in 1960

Let weekly papers in 1960 be x.

           7689 =  94% of  x

           7689 = (94/100)*x

           7689 * 100/94  = x

           8179.787      ≈  x
              
               x    ≈  8179.787

Number of weekly newspapers in 1960 rounded to the nearest whole number = 8180
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3 years ago
If the diameter of a circle pond is 42 m what is the area of the pond
svetoff [14.1K]

Answer:

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Step-by-step explanation:

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Area = 3.14 x 42² ➗ 4

Area = 5538.96/4

Area = 1384.74m²

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The millimeter marks are the small ones between the larger centimeter marks with numbers if using the centimeter side
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