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balu736 [363]
2 years ago
7

Find the midpoint of the segment with the following endpoints. (10,5) and (6,9)

Mathematics
2 answers:
deff fn [24]2 years ago
5 0
(8,7) i already did this question before
Nikitich [7]2 years ago
3 0

Answer:

(8 ,7 )

Step-by-step explanation:

Midpoint Formula: (\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )

Simply plug in your 2 coordinates into the midpoint formula to find midpoint:

(\frac{10+6}{2} ,\frac{5+9}{2} )

(\frac{16}{2} ,\frac{14}{2} )

(8 ,7 )

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$12.75 is 25% of $50
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Galina-37 [17]

mean = (sum of values) / (number of values)

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Which of the following is the quotient of the rational expression 4x/2x-1 divided by 3x+2/x+5
expeople1 [14]

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Step-by-step explanation:

7 0
2 years ago
AC = <br> Round your answer to the nearest hundredth.
dusya [7]

Answer:

AC ≈ 5.03

Step-by-step explanation:

We can solve the problem above using the trigonometric ratio, they are;

SOH CAH TOA

sin Ф = opposite / hypotenuse

cosФ= adjacent/ hypotenuse

tan Ф = opposite / adjacent

From the diagram above, in reference to angle B;

opposite =AC     and   adjacent =BC

Since we have opposite and adjacent, the best formula to use is

tanФ = opposite / adjacent

tan B = AC  / BC

tan 40 =  AC/ 6

Multiply both-side of the equation by 6

6× tan 40   =  AC/ 6  ×  6

At the right-hand side of the equation, 6 will cancel-out 6 leaving us with just AC

6×tan 40 = AC

5.034598  = AC

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4 0
3 years ago
Write an equation that represents the line.<br> Use exact numbers.<br> (-2,-1)<br> (4,6)
vazorg [7]

Check the picture below.

to get the equation of any straight line, we simply need two points off of it, so let's use those in the picture

(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{6})~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{6}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{(-2)}}} \implies \cfrac{6 -1}{4 +2} \implies \cfrac{ 5 }{ 6 }

\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{\cfrac{5}{6}}(x-\stackrel{x_1}{(-2)})\implies y-1=\cfrac{5}{6}(x+2) \\\\\\ y-1=\cfrac{5}{6}x+\cfrac{5}{3}\implies y=\cfrac{5}{6}x+\cfrac{5}{3}+1\implies y=\cfrac{5}{6}x+\cfrac{8}{3}

6 0
1 year ago
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