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Andrews [41]
3 years ago
13

Using sets what is the intersection of a=(-2-1,0,1,7) I=(-2,-1,1,2,8)​

Mathematics
1 answer:
polet [3.4K]3 years ago
4 0
The answer is (-2,-1,1)
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The answer would be 16.5
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Use the information provided to write the standard form eqaution of each parabola. y=2(x+2)^2+9
gtnhenbr [62]

Answer:

y = 2 x 2 + 8 x + 17

Step-by-step explanation:


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3 years ago
50 POINTS!!! jamie will choose between two catering companies for an upcoming party. company A chargers a set-up fee of $500 plu
Shtirlitz [24]

Answer:

60 guests

Step-by-step explanation:

We'll say that the number of guests is n.

We're looking for the n where both of these expressions are equal:

Company A (500 set-up fee, 25 for each guest)

500 + 25 * n

Company B (200 set-up fee, 30 for each guest)

200 + 30 * n

500 + 25n = 200 + 30n

Subtract 200 from both sides.

300 + 25n = 30n

Subtract 25n from both sides.

300 = 5n

We'll reverse the sides.

5n = 300

Divide both sides by 5.

n = 300/5 = 60

Thus, both companies are equal when n is 60, or when there are 60 guests.

5 0
3 years ago
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Choose the best coordinate system to find the volume of the portion of the solid sphere rho <_4 that lies between the cones φ
MrRissso [65]

Answer:

So,  the volume is:

\boxed{V=\frac{128\sqrt{2}\pi}{3}}

Step-by-step explanation:

We get the limits of integration:

R=\left\lbrace(\rho, \varphi, \theta):\, 0\leq \rho \leq  4,\, \frac{\pi}{4}\leq \varphi\leq \frac{3\pi}{4},\, 0\leq \theta \leq 2\pi\right\rbrace

We use the spherical coordinates and  we calculate a triple integral:

V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_0^4  \rho^2 \sin \varphi \, d\rho\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \left[\frac{\rho^3}{3}\right]_0^4\, d\varphi\, d\theta\\\\V=\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin \varphi \cdot \frac{64}{3} \, d\varphi\, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} [-\cos \varphi]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}  \, d\theta\\\\V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\

we get:

V=\frac{64}{3} \int_0^{2\pi} \sqrt{2} \, d\theta\\\\V=\frac{64\sqrt{2}}{3}\cdot[\theta]_0^{2\pi}\\\\V=\frac{128\sqrt{2}\pi}{3}

So,  the volume is:

\boxed{V=\frac{128\sqrt{2}\pi}{3}}

4 0
3 years ago
4.88 to nearest tenths
bazaltina [42]

Answer:

4.9

Step-by-step explanation:

4.88

The first after the period is the tenths place, it is rounded up to 4.9

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