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Yuliya22 [10]
2 years ago
10

What is the slope between the points (-2, 3) and (1, 0)

Mathematics
1 answer:
antiseptic1488 [7]2 years ago
7 0
You’re going to use the formula: y2-y1/x2-x1
plug that in: 0 - 3/-1 - 2 = -3/-3 or 1
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3(x−0.2)=1.8+x find x
bonufazy [111]

Answer:

x=6/5

Step-by-step explanation:

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3 years ago
Prove the Pythagorean Theorem using similar triangles. The Pythagorean Theorem states that in a right triangle, the sum of the s
aivan3 [116]
For the answer to the question above asking to p<span>rove the Pythagorean Theorem using similar triangles. The Pythagorean Theorem states that in a right triangle, 
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For example:  Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 3 inches and 4 inches.  
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8 0
3 years ago
Need help ASAP! PLSS
Lady bird [3.3K]

Answer:

i think u add them

Step-by-step explanation:

9+ 7+16=32

6 0
3 years ago
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \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}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
Con someone please help me :(( !!!!!!!!!
Bas_tet [7]

Answer:

The answer is A. 145

Step-by-step explanation:

Hope this helped :  )

4 0
3 years ago
Read 2 more answers
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