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2 years ago

Liz train for rock climbing 4 days a week

Mathematics

1 answer:

Goshia [24]2 years ago

7 0

7 times 4 =28 this is cuz there is seven days in a week and 7x4

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How do I find the rate of change when going from 4 then 2 then 8 then 14

Oksi-84 [34.3K]

Answer:

1.75

Step-by-step explanation: 14 divided by 8 is 1.75

5 0

3 years ago

(Will Give Brainliest) −2(4

mariarad [96]

Answer: 16

2^4

2*2*2*2

--> 16

3 0

2 years ago

Read 2 more answers

A cable company offers two movie plans. One plan costs $129 per month plus $5 per movie. The other plan costs $150 per month for

Svetradugi [14.3K]

Answer:
You can’t make it the same, but you can make it almost the same.

Maggie would have to rent 4 movies to make the plans’ cost almost the same.

Explanation:
$150-$129= 21
21 divided by 5= 4.2 (You can’t rent 4.2 movies, so we go with 4)
4x5=20+$129= $149 (Almost $150)

So, Maggie would have to rent 4 movies to make the costs almost the same.

Hope this is helpful!!
P.S Sorry if there was another way to make the costs equal, I’m not the best at math!
:D

4 0

3 years ago

Which best explains why the equation 7×+3=7×+3 has infinitely many solutions

madreJ [45]

Left and right side are exactly same, therefore "<span>the equation is true for any value of x.</span>"

Option B

4 0

4 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