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Hatshy [7]
3 years ago
15

Witch of the following is a true statement about a circle inscribed in a regular polygon

Mathematics
1 answer:
ss7ja [257]3 years ago
4 0
The area of the circle is equal to the area of the polygon
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I need help pleaseeeeee
aleksandr82 [10.1K]
The second one!!! For further help use cymath.com
7 0
2 years ago
If x / 5 = y, what is 2x?
tino4ka555 [31]
<u>x / 5  =  y</u>

Multiply each side by  5 :    x  =  5y

Multiply each side by  2 :   <em>2x  =  10y</em>
6 0
3 years ago
Consider the following function. f(x) = 16 − x2/3 Find f(−64) and f(64). f(−64) = f(64) = Find all values c in (−64, 64) such th
VARVARA [1.3K]

Answer:

This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64).

Step-by-step explanation:

The given function is

f(x)=16-\frac{x^2}{3}

To find f(-64), we substitute x=-64 into the function.

f(-64)=16-\frac{(-64)^2}{3}

f(-64)=16-\frac{4096}{3}

f(-64)=-\frac{4048}{3}

To find f(64), we substitute x=64 into the function.

f(64)=16-\frac{(64)^2}{3}

f(64)=16-\frac{4096}{3}

f(64)=-\frac{4048}{3}

To find f'(c), we must first find f'(x).

f'(x)=-\frac{2x}{3}

This implies that;

f'(c)=-\frac{2c}{3}

f'(c)=0

\Rightarrow -\frac{2c}{3}=0

\Rightarrow -\frac{2c}{3}\times -\frac{3}{2}=0\times -\frac{3}{2}

c=0

For this function to satisfy the Rolle's Theorem;

It must be continuous on [-64,64].

It must be differentiable  on (-64,64).

and

f(-64)=f(64).

All the hypotheses are met, hence this does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64) is the correct choice.

6 0
2 years ago
Read 2 more answers
You have 9 math coins and each coin is worth -25 points. How many points is that all together?
lesya [120]
The answer is -225.
9 times -25 is -225.
9*-25=-225
6 0
3 years ago
Read 2 more answers
In the isosceles triangle ABC, we have AB=AC=4. The altitude from B meets AC at H. If AH=3(HC) then determine BC.
Inga [223]

Answer:

BC=√7

Step-by-step explanation:

AC=4

AC=AH+HC

=3HC+HC

=4HC

HC=1/4AC=1/4×4=1

AH=3HC=3×1=3

BH⊥ AC

AB=AC=4

BC=\sqrt{AB^2-AH^2} =\sqrt{4^2-3^2} =\sqrt{16-9} =\sqrt{7}

5 0
2 years ago
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