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3 years ago

Step by step explanation!

Mathematics

2 answers:

gregori [183]3 years ago

7 0

Answer:

20 out every100 are both..

therefore 1 out of evey five ...

0.20

Afina-wow [57]3 years ago

5 0

Answer:

yhe picture is blur.

Step-by-step explanation:

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HELP!

emmasim [6.3K]

A) Divide the total height by the average height:

26.4 / 1.2 = 22 total boys.

B) Divide total feet by total time:

440 feet / 20 seconds = 22 feet per second.\

Because the submarine is diving the rate would become negative 22 feet per second.

7 0

3 years ago

What two decimals equivalent to 4.200

irina1246 [14]

it can be. 4.2 and. 4.20

6 0

3 years ago

Find "C" when "F" = 120.<br>Round to tenths.<br> C= 5F-160 /9

GenaCL600 [577]

Answer:

C = 48.9

Step-by-step explanation:

C= 5F-160 /9

C= (5F -160)/ 9

Put F =120

C = (5(120) - 160)/9

C= (600 - 160)/9

C = 440/9

C = 48.89≈ 48.9

7 0

4 years ago

What is the slope of y=7-3x

Anna [14]

The equation used above is y-intercept form.

ex. y = mx + b

Whatever is in front of the x is your slope.

Your slope is -3

7 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