Efgh is a parallelogram. find each measure

A cylindrical well has a radius of 8 feet and a height of 13 feet. What is the volume of the well? Use 3.14 for pi. Round your a

Murrr4er [49]

Answer:

using pi *r^2 *h

volume = 2614.14 cubic ft

3 0

2 years ago

Read 2 more answers

A 13-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on level ground 5 fee

Tanzania [10]

The answer is 12 because (a^2 + b^2 = c^2)

5^2 + b^2 = 13^2
25 + b^2 = 169

169- 25 = 144
$\sqrt{144}$ Sqaure root of 144 is 12 so If we plug that in..
25 + 144 = 169
5^2 + 12^2 =13^2
This all checks out
~Hope This helps

8 0

2 years ago

What is the algebraic expressions of 8-6k

Komok [63]

Eight decreased by six times the number k

The difference of eight and the product of six and the number k

Eight minus six times k

Six multiplied to k is less than 8

4 0

2 years ago

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$