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kherson [118]
3 years ago
15

In triangle ABC, angle A is equal to angle C, BA=x+20,CA=4x-30, and BC=3x+14. What is the length of BC?​

Mathematics
1 answer:
Anna [14]3 years ago
3 0

Answer:

BC = 23

Step-by-step explanation:

Since ∠ A = ∠ C then the triangle is isosceles with AB = BC, that is

3x + 14 = x + 20 ( subtract x from both sides )

2x + 14 = 20 ( subtract 14 from both sides )

2x = 6 ( divide both sides by 2 )

x = 3

Thus

BC = 3x + 14 = 3(3) + 14 = 9 + 14 = 23

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Given points A (1, 2/3), B (x, -4/5), and C (-1/2, 4) determine the value of x such that all three points are collinear
AlladinOne [14]

Answer:

x=\frac{83}{50}

Step-by-step explanation:

we know that

If the three points are collinear

then

m_A_B=m_A_C

we have

A (1, 2/3), B (x, -4/5), and C (-1/2, 4)

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

step 1

Find the slope AB

we have

A(1,\frac{2}{3}),B(x,-\frac{4}{5})

substitute in the formula

m_A_B=\frac{-\frac{4}{5}-\frac{2}{3}}{x-1}

m_A_B=\frac{\frac{-12-10}{15}}{x-1}

m_A_B=-\frac{22}{15(x-1)}

step 2

Find the slope AC

we have

A(1,\frac{2}{3}),C(-\frac{1}{2},4)

substitute in the formula

m_A_C=\frac{4-\frac{2}{3}}{-\frac{1}{2}-1}

m_A_C=\frac{\frac{10}{3}}{-\frac{3}{2}}

m_A_C=-\frac{20}{9}

step 3

Equate the slopes

m_A_B=m_A_C

-\frac{22}{15(x-1)}=-\frac{20}{9}

solve for x

15(x-1)20=22(9)

300x-300=198

300x=198+300

300x=498

x=\frac{498}{300}

simplify

x=\frac{83}{50}

8 0
3 years ago
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