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3 years ago

What is 6 cm and 7 mm converted to

Mathematics

1 answer:

Mama L [17]3 years ago

8 0

Answer:

converted to what?

Step-by-step explanation:

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In the past ten years, the population of a city decreased from 90,000 to 75,000 Find the percent decrease.

vfiekz [6]

The answer is 31.5%.

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3 years ago

Plz help me with this

8090 [49]

Answer:

8.216 put a line on top of the six because it's repeating

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

9(-6X-2) -7 (-10X-2)

Maksim231197 [3]

9(-6x-2)-7(-10x-2)
Simplify, by distributing the 9 and the -7
-54x-18+70x+14
Combine like terms
16x-4 is the simplified expression

Hope this helped!

3 0

3 years ago

Whats the answer to number 9?

agasfer [191]

Answer:

B and D

Step-by-step explanation:

Because only those two equations on the graphs are straight. and are increasing or decreasing in a constant pace.

4 0

3 years ago

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.

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

4 years ago