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natulia [17]
3 years ago
6

A médium-sized paper cone has a diameter of 8 centimeters and a height of 10 centimeters. What is the volume of the cone?

Mathematics
1 answer:
lubasha [3.4K]3 years ago
5 0

Answer:

40/3 pi

Step-by-step explanation:

Volume of a cone = 1/3 *pi*h*r^2 , where h = height, and r = radius.

diameter = 2 radius, so radius = 8/2 = 4

The height = 10, so plug these values into the equation:

1/3 * pi * 10 * 4 =

40/3 pi

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6 0
2 years ago
I really need help what's the answer??
lys-0071 [83]
I think you just have to add all of the totals up??
5 0
3 years ago
HELP ASAP I DONT WANNA FAIL
PtichkaEL [24]

Answer: the third one

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A^2(64a^2-3)/3=27-4a^2/4
frutty [35]
The answer is a = 3/4 = 0.75

First get rid of the paranthesis,
{a}^{2} (64 {a}^{2}  - 3) = 64 {a}^{4} -  3 {a}^{2}
Then set the denominators equal:
\frac{ 256 {a}^{4}  - 12 {a}^{2}}{12}  =  \frac{81 - 12 {a}^{2} }{12}
Then remove the denominators and solve:
256 {a}^{4}  - 12 {a}^{2}  = 81 - 12 {a}^{2}
Eliminate -12a^2 by adding 12a^2 to both sides:
256 {a}^{4}  = 81
Take the fourth root of them or take the square root twice:
\sqrt[4]{256 {a}^{4} }  =  \sqrt[4]{81}   \\  4a = 3
Divide both sides by 4:
a =  \frac{3}{4}  = 0.75
7 0
3 years ago
plain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = x2 − 2x x2 − 4 if x ≠ 2 1 if x =
ElenaW [278]

Answer with Step-by-step explanation:

We are given that

f(x)=\left\{\begin{matrix}\dfrac{x^2-2x}{x^2-4}&,if\ \ x\neq 2 \\ 1&,if\ \ x=2\end{matrix}\right.

We have to explain that why the function is discontinuous at x=2

We know that if function is continuous at x=a then LHL=RHL=f(a).

f(x)=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}

LHL=Left hand limit when x <2

Substitute x=2-h

where h is small positive value >0

\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2-h}{2-h+2}

\lim_{h\rightarrow 0}\frac{2-h}{4-h}=\frac{2}{4}=\frac{1}{2}

Right hand limit =RHL when x> 2

Substitute

x=2+h

\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2+h}{2+h+2}=\lim_{h\rightarrow 0}\frac{2+h}{4+h}

=\frac{2}{4}=\frac{1}{2}

LHL=RHL=\frac{1}{2}

f(2)=1

LHL=RHL\neq f(2)

Hence, function is discontinuous at x=2

4 0
3 years ago
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