All categories

2 years ago

A cone has a volume 12 pie cubic inches it’s height 4 inches what is the radius?

Mathematics

1 answer:

insens350 [35]2 years ago

3 0

Answer:

Step-by-step explanation:

The formula of a volume of a cone:

$V=\dfrac{1}{3}\pi r^2H$

r - radius

H - height

We have V = 12π in³ and H = 4 in. Substitute:

$12\pi=\dfrac{1}{3}\pi(r^2)(4)\qquad\text{multiply both sides by 3}\\\\36\pi=4\pi r^2\qquad\text{divide both sides by}\ 4\pi\\\\9=r^2\to r=\sqrt9\to r=3$

You might be interested in

The shelf is made of 3 pieces of wood, each 0.3 meter wide. There is also 0.05 meter of space between the shelf and the wall on

lutik1710 [3]

Since there are 2 sides to the shelf, there is 0.05*2=0.05+0.05=0.1 meters of space between the shelf and wall in total. Next, since each piece of wood is 0.3 meters wide and we have 3 of them, the total width of the wood is 0.3+0.3+<something>
Note that <width of wood>+<space between shelf and well> =width of closet
I challenge you to finish this on your own - good luck, and feel free to ask with any further questions!

8 0

3 years ago

Please Help! Need by today!

ra1l [238]

Answer:

sorry but no one answered it sorry good lucked

Step-by-step explanation:

4 0

1 year ago

Read 2 more answers

What equation represent 3x3/5 as a multiple of a unit fraction

Agata [3.3K]

The answer is 9/5 as 3×3 is 9 and you are dividing it by 5.

4 0

2 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

2 years ago

Leah wants to carpet a room that is 12 feet wide and 15 feet long. Carpeting is sold by the square yard. Find the length and wid

Zigmanuir [339]

Answer:

length: 5 yd

width: 4 yd

area: 20 yd^2

Step-by-step explanation:

1 yd = 3 ft

15 ft * (1 yd)/(3 ft) = 5 yd

12 ft * (1 yd)/(3 ft) = 4 yd

A = LW

A = 5 yd * 4 yd = 20 yd^2

7 0

3 years ago

Read 2 more answers