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Tatiana [17]
3 years ago
14

Can someone please help asap

Mathematics
2 answers:
kondor19780726 [428]3 years ago
3 0
8 √2
This is the answer for the problem
lesya [120]3 years ago
3 0

Answer:

8 \sqrt{2}

Step-by-step explanation:

8²+8²=(BC)²

64+64= 128

BC²=128

BC=

8 \sqrt{2}

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PLEASE HELP!!!
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Step-by-step explanation:

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7 0
2 years ago
What division problem can be represented using the number line?
kow [346]

Answer: The Answer is A

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
What is the best way to gather information for each of the scenarios?
topjm [15]

(a) The team captain wants to know what his teammates eats before a match

Team captain know by taking a survey of teammates

(b) The team captain wants to know who can score most goals against the best player

This can be known by observation of games. so its observational study

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4 0
3 years ago
Read 2 more answers
Please i need help with this one​
swat32

Answer:

Step-by-step explanation:

use formula

F = 1/2*sqrt[ 4* r^2 -CD^2 ]  

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4 0
2 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.

<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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