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2 years ago

How many 2 quarter hours in half hours

Mathematics

1 answer:

Oxana [17]2 years ago

8 0

Answer:

1 1/3

Step-by-step explanation:

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Help and I’ll give brainliest

Vaselesa [24]

Answer:

They purchased 11 Packets

Step-by-step explanation:

C=11p + 30

6 0

3 years ago

For a certain year, the combined revenues for Pepsi cola and coca-cola were 57 billion. If the revenue for Pepsi co was 13 billi

Scrat [10]

A/b*c=d/8(5m*k) thats the eqation

6 0

3 years ago

Read 2 more answers

Which property is shown below?

ElenaW [278]

Answer:

Step-by-step explanation:

5+2y+3z=5+3z+2yA

Intercambie los lados para que todos los términos de las variables estén en el lado izquierdo.

5+3z+2yA=5+2y+3z

Resta 5 en los dos lados.

3z+2yA=5+2y+3z−5

Resta 5 de 5 para obtener 0.

3z+2yA=2y+3z

Resta 3z en los dos lados.

2yA=2y+3z−3z

Combina 3z y −3z para obtener 0.

2yA=2y

Anula 2 en ambos lados.

yA=y

Divide los dos lados por y.

Al dividir por y, se deshace la multiplicación por y.

Divide y por y.

A=1

6 0

2 years ago

470.47 in expanded form

Aleksandr-060686 [28]

Four hundred seventy and forty seven hundredth

7 0

3 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

<u>Step-by-step explanation:</u>

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

<u>Proof LHS → RHS</u>

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