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3 years ago

Which best describes the range of the function f(x)=2(3)^x

Mathematics

2 answers:

vitfil [10]3 years ago

5 0

Answer:

y>0

Step-by-step explanation:

antiseptic1488 [7]3 years ago

5 0

Answer:

$f(x)>0$

Step-by-step explanation:

In the picture below, the graph of f(x) = 2(3)^x is shown. Since, the horizontal asymptote is 0, the range is all real numbers higher than 0.

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Dvinal [7]

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3 years ago

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Given bc=4 cd=11 and bd=10 list angles of triangle bcd least to greatest

Nataly_w [17]

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3 years ago

HELP IS THIS correct??? !!!

Allushta [10]

Answer:

Step-by-step explanation:

x y

y = - 3(-2) + 1 = 7 -2 7

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3 years ago

Find y as a function of x if y"' + 9y' = 0, y(0) = 6, y'(0) = 9, y"(0) = 18. y(x) =

Andreas93 [3]

Answer:

y(x) = 8 - 2cos 3x + 3sin 3x

Step-by-step explanation:

Given the equation:

y''' + 9y' = 0

The characteristic equation :

⇒r³ + 9r =0

Solving for r, we get r = 0 and r = ± 3i

The general equation for such equation is :

y = C₁ + C₂cos 3x + C₃sin 3x

Given:

y(0) = 6

Thus, Applying in the above equation, we get

6 = C₁ + C₂ .......................................................1

Differentiating y , we get:

y' = -3C₂sin 3x + 3C₃cos 3x

Given:

y'(0) = 9

Thus, Applying in the above equation, we get

9 = 3C₃

or,

C₃ = 3

Differentiating y' , we get:

y'' = -9C₂cos 3x - 9C₃sin 3x

Given:

y''(0) = 18

Thus, Applying in the above equation, we get

18 = -9C₂

or,

C₂ = -2

Applying in equation 1 , we get:

C₁ = 8

Thus,

y(x) = 8 - 2cos 3x + 3sin 3x

4 0

3 years ago

Can you please help me

asambeis [7]

C. Hope this helps:)

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3 years ago