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kogti [31]
3 years ago
12

Solve for X A) 5 B) 6 C) 7 D) 4

Mathematics
1 answer:
satela [25.4K]3 years ago
7 0

Answer:

A) 5

Step-by-step explanation:

ratio of big triangle to small is 9:5

45/9=5

5x/5=x

x=5

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What is 0.2857142857 rounded to the nearest whole percent? pls help me​
Grace [21]

29% would be the answer if you are rounding up

6 0
3 years ago
Mary month earnings can be given as R(x)=13x+155 where x equals the number of hours worked josies monthly are given as Q(x)=9x+3
tatuchka [14]

Answer:

45hours

Step-by-step explanation:

Given

Mary month earnings to be R(x)=13x+155

Josies monthly earning is given as Q(x)=9x+335

If Josie and Mary work to earn the same salary, then R(x) = Q(x)

13x+155 = 9x+335

subtract 155 from both sides

13x+155-155 = 9x+335-155

13x = 9x+180

subtract 9x from both sides

13x-9x = 9x+180-9x

4x = 180

x = 180/4

x = 45

Since  x equals the number of hours worked by Josie and Mary, this means that Josie and Mary must work for 45hours to earn the same salary

6 0
3 years ago
What is the area of the composite figure?<br><br> 70 cm2<br> 100 cm2<br> 105 cm2<br> 130 cm2
Hitman42 [59]

Answer: 100cm^2

Step-by-step explanation:

i took the test

3 0
3 years ago
Read 2 more answers
Katie’s goal is to read 6 books every 3 months. Based on this goal, how many months will it take Katie
Andrews [41]

Answer:

12

Step-by-step explanation:

24/6=4 4x3=12

4 0
3 years ago
If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt
MatroZZZ [7]
Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t
\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle

with 0\le t\le1. Then we find the differential:

\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle
\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt

with 0\le t\le1.

Here, the line integral is

\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle
=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt
=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt
=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt
=x_1y_2-x_2y_1

as required.
4 0
4 years ago
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