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3 years ago

How to find the volume of a cone? Height- 8 in. Diameter- 7 in.

Mathematics

1 answer:

mestny [16]3 years ago

4 0

The formula for the volume of a cone is V = $\pi r^{2} \frac{h}{3}$

So the answer is 102.63

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Find the quotient of 48a^4-16a^2-32/8a^2-8

Novay_Z [31]

We know that if we had
8/6=2/2 times 4/3=1 times 4/3=4/3
find common factors in top and obttom
factor
48a^4-16a^2-32=(16)(a-1)(a+1)(3a^2+2)
8a^2-8=8(a-1)(a+1)
so we have
$\frac{48a^4-16a^2-32}{8a^2-8} = \frac{(2)(8)(a-1)(a+1)(3a^2+2)}{(8)(a-1)(a+1)}= (\frac{8}{8}) (\frac{a-1}{a-1})( \frac{a+1}{a+1}) ( \frac{(2)(3a^{2}+2)}{1})=$ (2)(3a^2+2)=6a^2+4

answer is 6a^2+4

5 0

3 years ago

Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.

This requires

$\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)$

$\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\sin y+C$

So we have

$f(x,y)=-\cos x+\sin y+C$

which means $\vec F$ is indeed conservative. By the fundamental theorem, we have

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}$

3 0

3 years ago

Hey there! posted picture of question help please

Amiraneli [1.4K]

For this case we have the following function:
f (x) = (1/6) ^ x
We must evaluate the function for x = 3
We have then:
f (3) = (1/6) ^ 3
Rewriting:
f (3) = (1/216)
Answer:
The function evaluated at x = 3 is:
f (3) = (1/216)
option C

8 0

3 years ago

Read 2 more answers

Graph y = -1/4x + 6. Please help!!

OverLord2011 [107]

Answer:

6 1/4

Step-by-step explanation:

6 0

3 years ago

A normal probability distribution

julia-pushkina [17]

Answer:

Option B) is a continuous probability distribution

Step-by-step explanation:

Properties of a normal probability distribution:

The mean, mode and median of the data is same.
It is a continuous probability distribution.
The area under the curve is 1.
The probability that any random variable X at a particular value is zero.

Thus, from the properties of normal distribution, the correct answer is:

Option B) is a continuous probability distribution

6 0

3 years ago