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anzhelika [568]
3 years ago
14

20. Find the midpoint between the given points.

Mathematics
1 answer:
AlexFokin [52]3 years ago
5 0

Hi there! :)

Answer:

\huge\boxed{(-1, -10.5)}

Use the midpoint formula to solve for the midpoint:

(x_{m}, y_{m}) = (\frac{x_{1}+x_{2} }{2} , \frac{y_{1}+y_{2}}{2})

Plug in the points:

(x_{m}, y_{m}) = (\frac{-5+3 }{2} , \frac{-13 -8}{2})

Simplify:

(x_{m}, y_{m}) = (\frac{-2 }{2} , \frac{-21}{2})

(x_{m}, y_{m}) = (-1 , -10.5)

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Mark is going to an awards dinner and wants to dress appropriately. He had one blue shirt, one white dress shirt, one black dres
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Answer:

<u>All the four statements are true , which are -</u>

(A) The subset contains contains of all the outfits that have grey slacks .

(B)  The subset contains contains of all the outfits that have grey slacks and red tie .

(C) The subset contains contains of all the outfits that have grey slacks and blue shirt .

(D)  The subset contains contains of all the outfits that do not have black slacks .

Step-by-step explanation:

<u>We are given with Outfit 2 , 4 and 6</u>

Outfit 2 -   Blue     grey     red

Outfit 4 -   White    grey     red

Outfit 6   - Black     grey    red

(A) Since only 2,4, and 6 contain grey slacks, and no other outfit contains grey slacks, those are the subsets that contain grey slacks. Thus , this statement is true .

(B) Just outfits 2, 4, and 6 have grey slacks and a red tie, proving the statement. Other clothes do not include grey slacks and a red tie; they may include a red tie but not both grey slacks and a red tie. Hence , the statement is true .

(C) Since the outfit 2 has grey slacks and a blue shirt, the statement is correct; no other outfit has both grey slacks and a blue shirt. Therefore , this statement is also correct .

(D) The argument is correct since all of the subsets contain grey slacks and no black slacks.

Hence , all the options A , B , C , D are correct and describes the subset .

<u>The given question is incomplete , the complete question is attached here - </u>

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3 years ago
The volume of a? cone-shaped hole is 49 pi ft cubed . If the hole is 3 ft? deep, what is the radius of the? hole?
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\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ h=3\\ V=49\pi \end{cases}\implies 49\pi =\cfrac{\pi r^2(3)}{3}\implies 49\pi =\pi r^2 \\\\\\ \cfrac{49\pi }{\pi }=r^2\implies 49=r^2\implies \sqrt{49}=r\implies 7=r

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