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kari74 [83]
3 years ago
7

Please I really need the answer of this question!!!

Mathematics
1 answer:
julia-pushkina [17]3 years ago
8 0
2nd option is the correct answer 
sofia is correct.
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The blades of a windmill turn on an axis that is 35 feet above the ground. The blades are 10 feet long and complete two rotation
7nadin3 [17]
<span>It could be C. If the blade spends twice in a minute, it would be 30 seconds. 30=2pi/B=pi/15. </span>
6 0
3 years ago
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Zoe had a board 5 1/4 feet long she cut off a piece now the board is 3 5/6 feet long how long was the piece she cut off
11Alexandr11 [23.1K]

5 1/4-3 5/6=1 5/12

Hope this helped!!! :P

7 0
3 years ago
fundraiser the least popular item was chips which they sold 21 bags of.The chips cost a total of $48.10 were the chips profitabl
spin [16.1K]

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Step-by-step explanation:

3 0
3 years ago
How many eights are in 4 draw models
Sergeeva-Olga [200]
8 divided by 4 = 2

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6 0
3 years ago
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
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