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Troyanec [42]
3 years ago
12

What is the area of the circle when the diameter is 2

Mathematics
1 answer:
mel-nik [20]3 years ago
7 0

Answer:

3.14

Step-by-step explanation:

because its diameter that is 2, you divide it in half which should be one, that will be the radius. to get the area from the radius you multiply it by its self and multiply that by 3.14 which is the circumference of a circle.

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Help plz!!!!!!!!!!!!!
trapecia [35]

Answer:

i believe the relative max coordinate is (2,4)

relative min is (-2,0) and (4,0)

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6 0
3 years ago
Help please this is urgent
yulyashka [42]

Answer:

64 degrees.

Step-by-step explanation:

This is because the arrow that is not horizontal is pointing to that number.

3 0
3 years ago
Need help -5(1+2k)-8(-4+5k)
Nataliya [291]

Answer:

27 - 50k

Simplify

1. Distribute

-5 ( 1 + 2k ) - 8 ( -4 + 5k )

-5 - 10k - 8 ( -4 + 5k )

2. Distribute

-5 - 10k - 8 ( -4 + 5k )

-5 - 10k + 32 - 40k

3. Add the numbers

-5 - 10k + 32 - 40k

27 - 10k - 40k

4. Add the same term to both sides of the equation

27 - 10k - 40k

27 - 50k

6 0
3 years ago
Explain the difference between a greatest common factor and a least common factor
Simora [160]
Greatest Common Is The More Times It Can Go In To That Number. And The Least Is The Least Many Times.

4 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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