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3 years ago

Does anyone know how to find indicated values??? I am struggling!

1 answer:

5 0

To solve this problem, we must understand function notation and the operations that go along with it. For Example, in this problem, f(x)g(x) means that we must multiply f(x) by g(x). What this means is we should multiply the functions by one another, as modeled below:

f(x)g(x)

(3x^2 + 5x)(7x +3)

To simplify this expression, we must use the FOIL method for multiplying binomials, which means we multiply together the first, outer, inner, and last terms and then add them together. This is modeled below:

(3x^2)(7x) + (5x)(7x) + (3x^2)(3) + (5x)(3)

To simplify this expression, we must first perform the multiplication within each term:

21x^3 + 35x^2 + 9x^2 + 15x

Next, we can add together the two terms with the same exponents:

21x^3 + 44x^2 + 15x

Therefore, your answer is 21x^3 + 44x^2 + 15x.

Hope this helps!

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PLEASE HELP! Explain your answer, THANK YOU!

Misha Larkins [42]

Answer:

-2/-3

Step-by-step explanation:

(75, 60) (45, 40)

75 60 since 75 is larger than 45, we subtract = -30

45 40 since 60 is larger than 40, we subtract = -20

= -20/-30

reduce

= -2/-3

3 0

4 years ago

HELP PLEASEWhat is the value of the exponential expression below?

vladimir2022 [97]

Hi, I got C because you just put 1/2 in for an exponent.. so, 16 of an exponent of 1/2.. it would be letter C. 4

6 0

3 years ago

Read 2 more answers

An analyst in Bank of America, Santa Clara University Branch, believes that on Mondays customers arrive at the bank with rate 50

Aleksandr-060686 [28]

Answer:

0.0821

Step-by-step explanation:

If the arrival rate is 50 customers per hour (60 minutes), then the expected number of customers in 3 minutes is:

$\lambda=\frac{50}{60}*3\\ \lambda=2.5\ customers/3\ minutes$

Assuming a Poisson distribution with 2.5 customers per 3 minutes, the probability that zero customers arrive for 3 minutes is:

$P(x=k)=\frac{\lambda^k*e^{-\lambda}}{k!}\\P(x=0)=\frac{2.5^0*e^{-2.5}}{0!}\\ P(x=0)=0.0821$

The probability that the bank stays empty for three minutes is 0.0821.

6 0

4 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$