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2 years ago

Set up a proportion to solve for x in the following similar triangles.

Mathematics

1 answer:

Alexus [3.1K]2 years ago

3 0

Answer:

add then subtract i think

Step-by-step explanation:

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If L is parallel to M, find the values of x, y, and all 3 angles.

lesya [120]

Answer:

8x - 14 = 5x + 34 (alternate exterior angles)

8x - 5x = 34 + 14

3x = 48

x = 48/3

x = 16

5y + 16 + 8x - 14 = 180 (linear pair)

5y + 16 + 8(16) - 14 = 180

5y + 16 + 128 - 14 = 180

5y + 130 = 180

5y = 180 - 130

5y = 50

y = 50/5

y = 10

∠ (8x - 14) = 114 [(8 × 16 -14)=114]

∠ (5x + 34) = 114 [(5 × 16 +34)=114]

∠ (5y + 16) = 66 [(5 × 10 +16)=66]

hope this helps you!

4 0

2 years ago

(5 1/4 - 2 1/2) + (6 2/3 - 3 1/2)

Vaselesa [24]

5 11/12 is the answer

8 0

3 years ago

Simplify 8r - 3j + 9j - 2r

Shkiper50 [21]

Answer:

6j + 6r :)))))))))))))))))

3 0

2 years ago

Read 2 more answers

PLEASE HELP!!! WORTH 20 POINTS!!!!

dexar [7]

1/1 because if you’ve already take one or orange out of the box you would only have one left to select two from the same box. you started with 2/2 and took one out put 1/1.

3 0

2 years ago

What are the vertical and horizontal asymptotes for the function f(x)=<br> 3x2/x2-4

Alecsey [184]

Answer: f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.

Step-by-step explanation:

Given function: $f(x)=\dfrac{3x^2}{x^2-4}$

The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.

i.e. $x^2-4=0\Rightarrow\ x^2=4\Rightarrow\ x=\pm2$

Hence, f(x) will have vertical asymptotes at x=-2 and x=2.

To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.

So, the graph will horizontal asymptote at $y=\dfrac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}$

i.e. $y=\dfrac{3}{1}=3$

Hence, f(x) will have horizontal asymptote at y=3.

3 0

3 years ago