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3 years ago

If f(x) = 3^(x + 4), find f(2). A. 729 B. 243 C. 81 D. 18

Mathematics

1 answer:

Umnica [9.8K]3 years ago

4 0

Answer:

f(2) = 729

Step-by-step explanation:

f(x) = 3^(x + 4)

Let x =2

f(2) = 3^(2 + 4)

f(2) = 3^(6)

f(2) = 729

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Consider the following functions:

Nina [5.8K]

Answer:

Step-by-step explanation:

(f+g)(1) = f(1) +g(1) = -1 + 2 = 1

The pair (1, -1) in the definition of f tells you f(1) = -1.

The pair (1, 2) in the definition of g tells you g(1) = 2.

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3 years ago

F(x) 3x^4-4x^2-3<br> Is this function odd?

Karo-lina-s [1.5K]

Answer:

No.

Step-by-step explanation:

f(x) = 3x^4-4x^2-3

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Here f(-x) = 3(-x)^4 - 4(-x)^2 - 3

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2 years ago

Write the the right answer and how you got to the answer. BEST ANSWER GETS BRAINLIEST

Gala2k [10]

I split it into parts. The small rectangle sticking out is 2x3 which is a 6 and the big rectangle 4x8 which is 32. The triangle is 8x2/2 which is 8. 6+32+8 = 46

6 0

3 years ago

Read 2 more answers

The volume V of a cylinder is computed using the values 8.8m for the diameter and 5.8m for the height. Use the linear approximat

Delvig [45]

Answer:

The maximum error is approximately Ev=24%

Step-by-step explanation:

the volume of the cylinder V is

V= π/4*H*D²

where H= height and D= diameter

the variation of V will be

dV = (∂V/∂H)*dH + (∂V/∂D)*dD

dV = π/4*D²*dH +π/2*H*D*dD

if we divide by the volume V

dV /V = (π/4*D²*dH +π/2*H*D*dD )/( π/4*H*D²) = dH/H + 2*dD/D

dV /V = dH/H + 2*dD/D

then we can approximate

error in V= Ev= ΔV/V ≈ dV/V

error in H= Eh=ΔH/H ≈ dH/H

error in D= Ed=ΔD/D ≈ dD/D

thus

Ev= Eh + 2*Ed

since Ed=Eh=E=8%

Ev= Eh + 2*Ed =3*E=3*8%=24%

Ev= 24%

therefore the maximum error is approximately Ev=24%

5 0

3 years ago

the mgf of a random variable x is e^3(e^t-1). Find P[mean - standard deviation squared < X < 1/2( mean + standard deviatio

postnew [5]

The given MGF is that for a random variable following a Poisson distribution with parameter $\lambda=3$ .

This means $\mathbb E(X)=\mathbb V(X)=\lambda$ , and $X$ has PMF

$f_X(x)=\begin{cases}\dfrac{3^xe^{-3}}{x!}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}$

So, the desired probability is

$\mathbb P\left(\lambda-\lambda^2$

This is equivalent to

$\displaystyle\sum_{x=0}^2\mathbb P(X=x)=\sum_{x=0}^2\frac{3^x}{x!e^3}=\frac{17}{2e^3}\approx0.4232$

8 0

3 years ago