All categories

4 years ago

How many times does 4 go in to 82 evenly

Mathematics

1 answer:

vodomira [7]4 years ago

4 0

use division;

82 ÷ 4 = 20.5

so 4 goes into 82 only 20 times evenly.

hope that helps :D

You might be interested in

Solve for x. Round to the nearest tenth, if necessary. F х 72 8.8 G

Irina-Kira [14]

Answer:

2.8592933268

Step-by-step explanation:

tan(72) = 8.8/x

=> x = 8.8/tan(72) = 2.859293326849

3 0

3 years ago

The Alan Company bought 80 tickets for a jazz company. After giving away 20 tickets to customers, the company sold the rest to t

aleksandrvk [35]

Each employ paid 166.67 for each ticket

4 0

3 years ago

The water park had a total of 1,212 visitors on Friday, Saturday, and Sunday. If the same number of people visited each day, how

Archy [21]

The answer is 404 visitors per day, Because 3/1212=404

5 0

3 years ago

Read 2 more answers

Kia's math test scores are shown. In order to join the math club she must have at least an 80 test score. What is the minimum sc

Wewaii [24]

The answer would be 81

6 0

3 years ago

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 4, 1), (0, 4, 1), and back to the origin under

Shkiper50 [21]

The work is equal to the line integral of $\vec F$ over each line segment.

Parameterize the paths

from (0, 0, 0) to (2, 0, 0) by $\vec r_1(t)=t\,\vec\imath$ with $0\le t\le2$ ,
from (2, 0, 0) to (2, 4, 1) by $\vec r_2(t)=2\,\vec\imath+4t\,\vec\jmath+t\,\vec k$ with $0\le t\le1$ ,
from (2, 4, 1) to (0, 4, 1) by $\vec r_3(t)=(2-t)\,\vec\imath+4\,\vec\jmath+\vec k$ with $0\le t\le2$ , and
from (0, 4, 1) to (0, 0, 0) by $\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k$ with $0\le t\le1$

The work done by $\vec F$ over each segment (call them $C_1,\ldots,C_4$ ) is

$\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0$

$\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3$

$\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2$

$\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3$

Then the total work done by $\vec F$ over the particle's path is 46.

8 0

4 years ago