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Varvara68 [4.7K]
3 years ago
8

The volume of a cube is given by the expression 3 and its surface area is given by the expression 62, where s is the length of t

he cube's side. What are the volume and surface area of a cube with a side length of 2 inches?
Mathematics
1 answer:
Zigmanuir [339]3 years ago
8 0
V=s³
SA=6s²



sub 2 for s
V=2³=8 cubic inches
SA=6(2²)=6(4)=24 square inches
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Step-by-step explanation:

We are given Jada's distance in kilometers and Han's distance in meters. To compare the two we need to have them both in the same unit of measurement.

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Compared to Han, who ran 8500 meters, Jada ran farther. 15250>8500.

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If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,
julsineya [31]
<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)

(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0
3 years ago
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