All categories

2 years ago

How do i find the volume of a cone with a radius of 9ft and a height of 24ft

Mathematics

1 answer:

ycow [4]2 years ago

8 0

Answer: V≈2035.75ft³

That is a small 3 after ft

HOPE THIS HELPS

You might be interested in

Find the length of YZ

emmainna [20.7K]

Larger triangle’s base length

a^2 + b^2 = c^2
a^2 + 3^2 = 8^2
a^2 = 8^2 - (3^2)
sqrt(a^2) = sqrt(55)
a = sqrt(55)
__________________

Smaller triangle’s base length:

The same formula applies.
a^2 + 3^2 = 5^2
a^2 = 5^2 - (3^2)
sqrt(a^2) = sqrt(16)
a = 4
____________

The finale!

Add the two side lengths of a, which is sqrt(55) + 4 (exact answer)
or... 11.416 (unrounded to thousandths place)

Good luck to you!

5 0

3 years ago

What is the solution set for the molar concentration of

malfutka [58]

Answer: B

Step-by-step explanation:

I got it right on Edge

4 0

3 years ago

Read 2 more answers

Find the Perimeter and Area

igor_vitrenko [27]

Answer:

Area = 20.28 cm^2

Perimeter = 16.28 cm

Step-by-step explanation:

The figure is composed of a half circle and a rectangle,

The circle has a diameter of 7 - 2 - 1 = 4cm and the rectangle is of side 2 cm and 7 cm.

Therefore the area is :

Area = A rectangle + Acircle/2 = 2*7 + pi*r^2/2 = 20.28 cm^2

The perimeter is :

Perimiter = 1 + 2 + 7 + Pcircle/2 = 1 + 2 + 7 + 2*pi*r/2 = 16.28 cm

3 0

2 years ago

Two numbers that multiply to -100 and add to 15

Whitepunk [10]

Answer:

-5 and 20

Step-by-step explanation:

-5 + 20 = 15

-5 * 20 = -100

3 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

How do i find the volume of a cone with a radius of 9ft and a height of 24ft​

How do i find the volume of a cone with a radius of 9ft and a height of 24ft