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Sergeu [11.5K]
2 years ago
5

16. Use the vertical line test to determine whether the relation is a function.

Mathematics
1 answer:
drek231 [11]2 years ago
7 0

Answer:

neither of them are

Step-by-step explanation:

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Rearranging formulae:
bazaltina [42]

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1. d/a+c=d

2. (m+21)/5=n

3. (1/2+2q)*4=p or 2+8q=p

4. (p-2a)/2pi=r

5. {[(5c+1)/2]+c}/3=a

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3 years ago
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How many times dose 12 go into 41
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3.42

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41/12 is 3.42

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It takes 8 men 96 days to build a tower. working at same rate, how long will it take 12 men
Kryger [21]

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dam gurll, how many questions u gon' ask

3 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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