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3 years ago

According to astronomers what is a light year

1 answer:

6 0

A unit of astronomical distance equivalent to the distance that light travels in one year, which is 9.4607 times 1012 km nearly 6 trillion miles.

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Solving the following system using the elimination of substitution method. 5a+3b=11 -2a+3=b

PSYCHO15rus [73]

Answer:

a = -2

b = 7

Step-by-step explanation:

5a+3b=11

-2a+3=b

substitute for b:

5a + 3(-2a + 3) = 11

5a -6a + 9 = 11

-a = 2

a = -2

b = 7

4 0

3 years ago

Help with 1-5

jekas [21]

it would be da families with 1 child

6 0

3 years ago

What is 36.5625 in fraction form

Romashka [77]

The answer is one hundred seventeen over three hundred twenty or 117/320

4 0

3 years ago

Plz help for exam!!! Solve for x and y.

ad-work [718]

Answer:

A x-198,y-99 $\sqrt{3}$

Step-by-step explanation:

since sin p/h you can say

perpendicular is 99

and hypotenuse is x

sin 30=99/x

x=198

and cos is b/h

base is y

and hypotenuse is x

cos 30=y/198

y-99 $\sqrt{3}$

6 0

3 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago