All categories

2 years ago

A cylinder has a base radius of 6ft and a height of 18ft. What is its volume in cubic ft, round to the nearest tenth?

Mathematics

1 answer:

Nookie1986 [14]2 years ago

3 0

Answer:2034.7

Step-by-step explanation:

The equation is pi •r^2•h

R=6

H=18

You might be interested in

2(x-1) ≥ 10 or 3-4x > 15<br> Solve the compound inequality, and show work!

Veseljchak [2.6K]

Answer:

$x\geq 6\text{ or } x$

Step-by-step explanation:

We have the compound inequality:

$2(x-1)\geq10\text{ or } 3-4x>15$

Let's solve each of them individually first:

We have:

$2(x-1)\geq10$

Divide both sides by 2:

$x-1\geq5$

Add 1 to both sides:

$x\geq6$

We have:

$3-4x>15$

Subtract from both sides:

$-4x>12$

Divide both sides by -4:

$x$

Hence, our solution set is:

$x\geq 6\text{ or } x$

5 0

2 years ago

Read 2 more answers

Find the common ratio and the seventh term of the following sequence:<br> ş, ğ, 2, 6, 18, ...

Karolina [17]

54 is the one after 18

4 0

2 years ago

What is the least common denominator of 6/17 and 5/18

Minchanka [31]

Answer:

The least common denominator is 306.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

3 years ago

Evaluate x6 + y(-9)<br> When x= -2, y = 4

11111nata11111 [884]

Answer:

the answer is -48. _______

4 0

2 years ago