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ELEN [110]
3 years ago
7

Javier​'s team scores an average of 15.5 points in 4 basketball games. Rhea​'s team scores an average of 29.5 points in the same

4 games. Use estimation to find how many times as many points Rhea​'s team scores compared to Javier​'s team. I need help plz
Mathematics
1 answer:
Dmitry [639]3 years ago
5 0

Answer:

well if you subtract 29.5 minus 15.5 youn wil get 14

Step-by-step explanation:

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Find the rate of change of total​ revenue, cost, and profit with respect to time. Assume that​ R(x) and​ C(x) are in dollars. ​R
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Answer:

Step-by-step explanation:

Given the Total revenue R(x) = 2x

Cost C(x) = 0.01x²+0.3x+30 where;

x = 30 and dx/dt = 9units per day.

Rate of change of revenue dR/dt = dR/dx • dx/dt

dR/dt = 2dx/dt

dR/dt = 2(9) = $18

Rate of change of revenue with respect to time is 18dollars/day.

Rate of change of cost dC/dt = dC/dx • dx/dt

dC/dt = (0.02x+0.3)dx/dt

dC/dt at x = 30 and dx/dt = 9 will give;

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4 0
3 years ago
What is 1 1/8 as a decmia
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8 0
3 years ago
Inverse laplace of [(1/s^2)-(48/s^5)]
Katen [24]
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
7 0
3 years ago
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