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olya-2409 [2.1K]
4 years ago
10

HELP!!Find the eighth term of the sequence given by the rule

Mathematics
1 answer:
murzikaleks [220]4 years ago
6 0

When you plug in 8 for n you get -14

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On average, a gym claims that 85% of its members renew their memberships every year. If the gym has 400 members, predict how man
lana [24]

Answer:

340 people would renew their membership if the gym had 400 members

The gym's claim is not accurate, because the prediction should be 340 people instead of 330

Step-by-step explanation:

With 400 members, we can predict how many will renew their membership by multiplying 400 by 0.85:

400(0.85)

= 340

So, this would predict that 340 people would renew their membership.

The gym's claim that 330 members would renew their memberships is not accurate, because the correct prediction should be 340 people.

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3 years ago
Jack typed 45 words in 1 minute. how many minutes will it take jack to type 990 words
Wittaler [7]

Answer:

22 minutes

Step-by-step explanation:

990/45=22

8 0
3 years ago
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1 foot = 12 inches

 divide 69 by 12

69/12 = 5.75 feet

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Kenny wrote the equation for a linear relationship shown below
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Y would equal -17! Good luck!
4 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
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