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3 years ago

Solve the equation. Check your solution.4v÷27=16

Mathematics

2 answers:

denis-greek [22]3 years ago

4 0

4v÷27=16
4v=16×27
4v=432
4=432÷4
v=108

svet-max [94.6K]3 years ago

3 0

V=108 because 4 ÷ 16 = 4 then you have v ÷ 27 = 4 and 27 times for is 108

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Alexeev081 [22]

Answer:

108º

Step-by-step explanation:

to find the sum of interior angles: ( n − 2 ) × 180 = x

x = 540º

n = # of interior angles

first, solve for n

(n - 2) • 180 = 540 distribute to 180 to (n - 2)

180n - 360 = 540 add 360 to both sides

180n = 900 divide both sides by 180

n = 5

to find measure of each angle:

540 ÷ 5 = 108º

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Read 2 more answers

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 $\checkmark$

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3 years ago