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2 years ago

If G(x) = (x + 5)2, and F(x) = x2 + 9x, what is the function G(F(x))?

1 answer:

6 0

Hello,

f(x)=x²+9x
g(x)=(x+5)²

g(f(x))=g(x²+9x)=((x²+9x)+5)²=(x²+9x+5)²=x^4+81x²+25+2*x²*9x +2*x²*5 +2*9x*5
=x^4+18x^3+91x²+90x+25

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Which ordered pair does not represent a point on the square?

Vikentia [17]

Answer:

Step-by-step explanation:

-2, 1 is not a point in the square but - 2, 2 is so the answer should be D -2,1

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2 years ago

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Solve the following ODE's: c) y* - 9y' + 18y = t^2

Nastasia [14]

Answer:

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = $C_1e^{3t}+C_2e^{6t}$

now calculating P.I.

$P.I. = \frac{t^2}{D^2 - 9D +18}$

$P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

hence the complete solution

y = C.F. + P.I.

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

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2 years ago

Assume that the Poisson distribution applies and that the mean number of hurricanes in a certain area is 6.9 per year. a. Find t

VLD [36.1K]

Answer:

a) 9.52% probability that, in a year, there will be 4 hurricanes.

b) 4.284 years are expected to have 4 hurricanes.

c) The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

6.9 per year.

This means that $\mu = 6.9$

a. Find the probability that, in a year, there will be 4 hurricanes.

This is P(X = 4).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-6.9}*(6.9)^{4}}{(4)!}$

$P(X = 4) = 0.0952$

9.52% probability that, in a year, there will be 4 hurricanes.

b. In a 45-year period, how many years are expected to have 4 hurricanes?

For each year, the probability is 0.0952.

Multiplying by 45

45*0.0952 = 4.284.

4.284 years are expected to have 4 hurricanes.

c. How does the result from part (b) compare to a recent period of 45 years in which 4 years had 4 hurricanes? Does the Poisson distribution work well here?

The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

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2 years ago

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Answer:

fixed

Step-by-step explanation:

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2 years ago

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Hello, may I please have some help on these 4 questions? Thanks very much! {Also, if you could answer them quick, that would be

wlad13 [49]

1.

Parallelogram with base a = 20 in. and height h = 16 in.
Triangle with base a = 20 in. and height b = 8 in.

so:

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$A=A_T+A_\Delta=\frac{1}{2}\cdot(b_1+b_2)\cdot h+\frac{1}{2}\cdot b_1\cdot x=\\\\=
\frac{1}{2}\cdot(14+4)\cdot 10+\frac{1}{2}\cdot 14\cdot8=\frac{1}{2}\cdot18\cdot 10+\frac{1}{2}\cdot 14\cdot8=90+56=\\\\=\boxed{146\text{ cm}^2}$

3.

We have three rectangles:

$A=A_1+A_2+A_3=1\cdot 4+6\cdot5+18\cdot6=4+30+108=\boxed{142\text{ ft}^2}$

4.

Area of a circle:

$A_\circ=\pi\cdot r^2=3.14\cdot5^2=3.14\cdot25=78.5 \text{ cm}^2$

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Area of the shaded region:

$A=A_\circ-A_R=78.5-48=\boxed{30.5\text{ cm}^2}$

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2 years ago