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Anna [14]
3 years ago
13

50 points and brainliest!

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
7 0

Answer:

3a-b/(x-b)=1/(x+1)Doing criss cross multiplication

3a(x+1)-b(x+1)=x-b

3ax+3a-bx-b=x-b

3ax-bx-x=b-3a-b

x(3a-b-1)=-3a

x=3a/(3a-b-1)

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MATH QUESTION!! WILL MARK BRAINLIEST!! PAST DUE HELP ASAP!!!
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Answer:

<em>Proof below</em>

Step-by-step explanation:

<u>Right Triangles</u>

In any right triangle, i.e., where one of its internal angles is 90°, some interesting relations stand. One of the most-used is Pythagora's Theorem.

In a right triangle with shorter sides a and b, and longest side c, called the hypotenuse, the following equation is satisfied:

c^2=a^2+b^2

The image provided in the question shows a line passing through points A(0,4) and B(3,0) that forms a right triangle with both axes.

The origin is marked as C(0,0) and the point M is the midpoint of the segment AB. We have to prove.

CM=\frac{1}{2}AB

First, find the coordinates of the midpoint M(xm,ym):

\displaystyle x_m=\frac{0+3}{2}=1.5

\displaystyle y_m=\frac{4+0}{2}=2

Thus, the midpoint is M( 1.5 , 2 )

Calculate the distance CM:

CM=\sqrt{(1.5-0)^2+(2-0)^2}

CM=\sqrt{2.25+4}=\sqrt{6.25}=2.5

CM=2.5

Now find the distance AB:

AB=\sqrt{(3-0)^2+(4-0)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}

AB=5

AB/2=2.5

It's proven CM is half of AB

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