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igor_vitrenko [27]
3 years ago
12

The product of 5 and a number n is at least 25

Mathematics
1 answer:
Crazy boy [7]3 years ago
4 0
5n≥25
and if you divide 5 on both sides, n≥5 (I don't know if you are supposed to solve or simply put it in the equation form).

Product refers to multiplication. so 5 and n are being multiplied in the equation.
However, instead of using an equal sign, the sign for "greater than or equal to" must be used. "At least" is a term equivalent to "greater than or equal to".
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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height
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Answer:

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Step-by-step explanation:

We are given he following information in the question:

y(t)=43t-23t^{2}

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\displaystyle\frac{\text{Displacement}}{\text{Time}}

1) 0.01

y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}

2) 0.005 s

y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}

3) 0.002 s

y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}

4) 0.001 s

y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}

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3 years ago
Simon put $15000 in an account with a simple interest rate of 4.5%. How much interest will be earned after 8 years?
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