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Neko [114]
3 years ago
10

The endpoints of CD are C(–8, 4) and D(6, –6). What are the coordinates of point P on CD such that P is the length of the line s

egment from D? (–2.75, 0.25) (–5, 2.5) (0.75, –2.25) (3.75, –3.75)

Mathematics
2 answers:
GalinKa [24]3 years ago
5 0
If you graph the end points C and D then graph the 4 points at the end it is difficult to tell which points are on CD without a line.  
Using the endpoints find the slope (change in y/ change in x) then substitute a point in to find the intercept.  
Slope = (-6-4)/(6- -8) = -5/7
Intercept equation (-6) = -5/7 (6) + b
b = -1.71428571429
Graphing the line shows only 2 points on the line (–2.75, 0.25) and <span>(0.75, –2.25)
I am confused by the part, "</span><span>P is the length of the line segment from D".  Were you given a length P to help you determine which point.  Using the distance formula to find the length from each point to D doesn't help determine which one is best with the information you have given.  The image shows the distances I calculated and the graphed points. 
I hope this helps!</span>

Mariulka [41]3 years ago
4 0

the answer is (–2.75, 0.25). hope this helps :)

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Nina [5.8K]

Answer:

The cost of cone of popcorn is $ 2.59.

Step-by-step explanation:

The tub has a volume of 540.473 and costs $10. The cone has a volume of 140.035.

The cost of per unit volume is

$10/540.473

The cost of cone of popcorn is

\frac{10}{540.473}\times 140.035\\\\2.59

7 0
3 years ago
784x44=? Step by step.
Triss [41]

Answer:

34496

Step-by-step explanation:

i dont know how 2 explain this but you have to do the traditional way of multiplying PLS MARK BRAINLIEST

7 0
2 years ago
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
1-12 I don’t know if you can please help me thanks
katovenus [111]

1 24

2 .41

3 180

4  2000

5   46

6  .18

7   60.5666

8   6.5

9   128

10   .8

11    3.628739

12   4000


pretty sure juh looked it up hope it helps!

4 0
3 years ago
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Lerok [7]

Answer:

Step-by-step explanation:

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Width = 2 1/2inch = 5/2inch

Area = l × width

So: 84 = l × 5/2

168 = 5l

l =>168/5

l = 33 3/5

7 0
3 years ago
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