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Masja [62]
3 years ago
15

For the people who helped me with the question

Mathematics
1 answer:
erik [133]3 years ago
4 0

Answer:

glad my answer helped you.

Step-by-step explanation:

hope that because i showed the steps, it will help you if you come across a problem like that in the future too!

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Aaliyah is a salesperson who sells computers at an electronics store. She makes a base
hjlf

Answer:

P=1.25x+70

Step-by-step explanation:

So we know that Aaliyah earns $1.25 commission for every computer she sells. No matter what, she will make $70 per day, so that would be our constant. The $1.25 would be our independent variable since it can change depending on the amount of computers x she sells. Therefore, our equation is:

P=1.25x+70

P represents the total pay, 1.25x represents the total commission for x computers, and the 70 is the base pay.

And we're done!

6 0
3 years ago
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
3 years ago
Pleaseee help with this! Giving 60 points!
Sonja [21]

Answer:

X = 2, -8

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
I’m struggling to answer this question on my test, can someone explain it to me?
Flauer [41]
Correct answer is Option A: 18/48 = 21/56

You’ll have to divide the numerator by the denominator of each fraction to see whether they have the same quotient:

18/48 = 0.375
21/56 = 0.375


The other options are not proportional.

B) 20/28 = 15/18:

20/28 = 0.714
15/18 = 0.833

Thus, 20/28 ≠ 15/18

C). 3/5 = 13/10
3/5 = 0.6
13/10 = 1.3

Hence, 3/5 ≠ 13/10

D) 7/10 = 10/7
7/10 = 0.7
10/7 = 1.43

Thus, 7/10 ≠ 10/7

These solutions prove that the correct answer is A) 18/48 = 21/56

Please mark my answers as the Briainliest if you find my explanations helpful :)
4 0
3 years ago
Reid at of the pumpkin pie. Vince ate 3 of the same pie. How much of the pie was left after Reid and Vince ate their pieces?
-Dominant- [34]
Whats the question? It doesn't make since
8 0
3 years ago
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