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sashaice [31]
3 years ago
13

How do u solve y pluse 4 pluse 3(y pluse 2)

Mathematics
2 answers:
larisa86 [58]3 years ago
8 0
y+4+3(y+2)

y+4+3y+6

y+3y+6+4

Answer:4y+10
lesya [120]3 years ago
3 0
Y + 4 + 3(y+2)    first distribute the 3
y + 4 + 3y + 6     then add like terms
4y + 10

therefore your answer should be  4y + 10
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Questions Below. Would Appreciate Help!
kherson [118]

Answer:

The function that could be the function described is;

f(x) = -10 \cdot cos \left (\dfrac{2 \cdot \pi }{3} \cdot x \right ) + 10

Step-by-step explanation:

The given parameters of the cosine function are;

The period of the cosine function = 3

The maximum value of the cosine function = 20

The minimum value of the cosine function = 0

The general form of the cosine function is presented as follows;

y = A·cos(ω·x - ∅) + k

Where;

\left | A \right | = The amplitude = Constant

The period, T = 2·π/ω

The phase shift, = ∅/ω

k = The vertical translation = Constant

Therefore, by comparison, we have;

T = 3 = 2·π/ω

∴ ω = 2·π/3

The range of value of the cosine of an angle are;

-1 ≤ cos(θ) ≤ 1

Therefore, when A = 10, cos(ω·x - ∅) = 1 (maximum value of cos(θ)) and k = 10, we have;

y = A × cos(ω·x - ∅) + k

y = 10 × 1 + 10 = 20 = The maximum value of the function

Similarly, when A = 10, cos(ω·x - ∅) = -1 (minimum value of cos(θ)) and k = 10, we get;

y = 10 × -1 + 10 = 0 = The minimum value of the function

Given that the function is a reflection of the parent function, we can have;

A = -10, cos(ω·x - ∅) = -1 (minimum value of cos(θ)) and k = 10, to get;

y = -10 × -1 + 10 = 20 = The maximum value of the function

Similarly, for cos(ω·x - ∅) = 1 we get;

y = -10 × 1 + 10 = 0 = The minimum value of the function

Therefore, the likely values of the function are therefore;

A = -10, k = 10

The function is therefore presented as follows;

y = -10 × cos(2·π/3·x) + 10

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