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3 years ago

Favorite you tuber?I like watching kendallgray and Demoranch and all them

Mathematics

1 answer:

Akimi4 [234]3 years ago

4 0

Prince Family. . . . .

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What is the perimeter of 2 1/2 and 5 1/3

Usimov [2.4K]

I am so sorry this is the wrong answer

8 0

3 years ago

55. A triangle has two angles measuring 46 and 14. Determine the measure of the third angle.

Step2247 [10]

Answer:

1. 180-(46+14)=120

2.46+14+120=180

Step-by-step explanation:

4 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

Y=6x-11

Firlakuza [10]

Elimination is the right answer

3 0

2 years ago

Read 2 more answers

Shelley drove from New Haven, Connecticut, to New York City in 90 minutes. Which equation relates the distance she traveled to h

lisov135 [29]

Answer:

Step-by-step explanation:

d = r * t

None of them are really correct, but the closest one is B. Look at the equation. Rate is multiplied by time. The result is distance.

The 90 minutes should be changed to hours. No speedometer on a car is rated in feet per second.

3 0

3 years ago

Favorite you tuber?I like watching kendallgray and Demoranch and all them ​

Favorite you tuber?I like watching kendallgray and Demoranch and all them