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Triss [41]
3 years ago
14

Answer these two question  ASAP

Mathematics
2 answers:
prohojiy [21]3 years ago
6 0

Answer:

explain further on this pls?



Evgen [1.6K]3 years ago
5 0

Answer:

the answer would be C) square root of 250 units

Step-by-step explanation:

5x5=25 and 15x15=225 you would then add them together and get 250 and the reason why it isnt 250 units is  because to find the hypotenuse you would need to find the square root of 250! Hope this helps! :3

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Wha it is 2,200 divided by 30 and 2,300 divided by 30
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Answer:

2,200/30=73.3333333

2,300/30=76.6666667

Step-by-step explanation:

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What is the area of the semicircle (use 3.14 for p)?<br> Please help! Will mark brainlyest.
Marizza181 [45]

Answer:

127.17 cm²

Step-by-step explanation:

  • Area of a semicircle: 1/2*π*r²
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A = 1/2*π*r²

A = 1/2*3.14*9²

A = 1/2*3.14*81

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A man can drive a motorboat 70 miles down the Colorado River in the same amount of time that he can drive 40 miles upstream. Fin
pochemuha

The speed of the current is 40.34 mph approximately.

<u>SOLUTION: </u>

Given, a man can drive a motorboat 70 miles down the Colorado River in the same amount of time that he can drive 40 miles upstream.  

We have to find the speed of the current if the speed of the boat is 11 mph in still water. Now, let the speed of river be a mph.  Then, speed of boat in upstream will be a-11 mph and speed in downstream will be a+11 mph.

And, we know that, \text{ distance } =\text{ speed }\times \text{ time }

\begin{array}{l}{\text { So, for upstream } \rightarrow 40=(a-11) \times \text { time taken } \rightarrow \text { time taken }=\frac{40}{a-11}} \\\\ {\text { And for downstream } \rightarrow 70=(a+11) \times \text { time taken } \rightarrow \text { time taken }=\frac{70}{a+11}}\end{array}

We are given that, time taken for both are same. So \frac{40}{a-11}=\frac{70}{a+11}

\begin{array}{l}{\rightarrow 40(a+11)=70(a-11)} \\\\ {\rightarrow 40 a+440=70 a-770} \\\\ {\rightarrow 70 a-40 a=770+440} \\\\ {\rightarrow 30 a=1210} \\\\ {\rightarrow a=40.33}\end{array}

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