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3 years ago

Name the following segment or point. Given: L, M, N are midpoints median to segment BL segment AS segment AM.

Mathematics

1 answer:

AnnyKZ [126]3 years ago

8 0

Check the attached file for the answer.

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Simplify the expression |-0.8 X 10|

eduard

I think -8 but i’m not sure tho

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3 years ago

Read 2 more answers

Help me pls it’s urgent

earnstyle [38]

Answer:

<A = 115

<C = 50

Step-by-step explanation:

<A + <B + <C = 180

(4X -13) +15 +(X+18) =180

5X +20 =180

5X = 160

X= 32

<C = 32+18 = 50

3 0

3 years ago

I know this question is easy, but I don’t really know how to do it? Help? If possible put a quick explanation, it’s fine if not!

lorasvet [3.4K]

Answer:

She left $10.62 the total was $58.87

Hope this helps :)

3 0

3 years ago

klio [65]

Sqrt of 125
10^2 +b^2= 15^2
100 +b^2=225
B^2=125
No perfect square for 125 so leave it in radical form

3 0

3 years ago

In a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft. Find: Chapter Reference b The length of the

Butoxors [25]

Answer:

The length of the angle bisector of angle ∠A is 6.01.

Step-by-step explanation:

It is given that length of leg AC = 5 ft and the hypotenuse AB = 13 ft.

Using pythagoras theorem

$(AB)^2=(BC)^2+(AC)^2$

$(13)^2=(BC)^2+(5)^2$

$169=(BC)^2+25$

$BC=12$

$\sin A=\frac{\text{perpendicular}}{\text{hypotenuse}}$

$\sin A=\frac{BC}{AB}$

$A=\sin ^{-1}\frac{12}{13}$

$A=67.38$

Bisector divides the angle in two equal parts, therefore,

$A'=\frac{67.38}{2} =33.69$

In triangle ACD.

$\cos A'=\frac{\text{Base}}{\text{Hypotenuse}}$

$\cos A'=\frac{AC}{AD}$

$\cos (33.69^{\circ})=\frac{5}{AD}$

$0.832=\frac{5}{AD}$

$AD=\frac{5}{0.832} =6.009\approx 6.01$

Therefore the length of the angle bisector of angle ∠A is 6.01.

4 0

3 years ago