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Suppose that the functions g and h are defined for all real numbers x as follows. gx = x − 3x
hx = 5x + 2
Write the expressions for (g - h)(x) and (g * h)(x) and evaluate (g + h)(−2).

Answers

Answer 1
Answer:

Answer:

Step-by-step explanation:

Given the functions g(x) = x − 3x  and h(x) = 5x + 2, we are to calculatae for the expression;

a) (g - h)(x)  an (g * h)(x)

(g - h)(x)  = g(x) - h(x)

(g - h)(x)  = x − 3x -(5x+2)

(g-h)(x) = x-3x-5x-2

(g-h)(x) =-7x-2

b)  (g * h)(x) =  g(x) * h(x)

 (g * h)(x)  = (x − 3x)(5x+2)

(g * h)(x) = 5x²+2x-15x²-6x

(g * h)(x) = 5x²-15x²+2x-6x

(g * h)(x) = -10x²-4x

c) To get (g + h)(−2), we need to first calculate (g + h)(x) as shown;

 (g + h)(x)  an (g * h)(x)

(g + h)(x)  = g(x) +h(x)

(g + h)(x)  = x − 3x + (5x+2)

(g+h)(x) = x-3x+5x+2

(g+h)(x) =3x+2

Substituting x = -2 into the resulting function;

(g+h)(-2) = 3(-2)+2

(g+h)(-2) = -6+2

(g+h)(-2) = -4

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Vincent wrote an example of a proportion on the board.Evaluate Vincent’s proportion to determine if it is correct. Explain your reasoning.

Answers

Answer:

Vincent’s proportion is incorrect. His corresponding parts are not in the same position. The heights and bases are in different positions.

Step-by-step explanation:

I got it right on my assignment

Answer:    Vincent’s proportion is incorrect. His corresponding parts are not in the same position. The heights and bases are in different positions.

Step-by-step explanation: i got it right on my assignment

Please help 30 points plus brainlyist who does firstDetermine which expressions can be simplified further, and which cannot. Sort the expressions into the correct
category
2x + 3y
Can Be Simplified
Cannot Be Simplified
x + x
4r +
7y + 1
4y + 4x
y + 2y

Answers

Answer:

can - y +2y

9x+6x

4x+x

can't 4y+4x

7y+1

2x+3y

Answer:

Step-by-step explanation:

expresion can be simplified is they have like terms such as

x+x=2x

y+2y=3y

expresions can NOT be simplified if they have difrerent variables or just one number suchh as

2x+3y

7y+1

4y+4x

I do not know what is 4r+

20% of the 35 vendors at a craft show sell jewelry. How many of the vendors sell jewelry? Please HURRYYYY

Answers

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What is the sum of 52.38 divided by 16

Answers

52.38 divided by 16 is 3.27

Find two consecutive positive integers such that the square of the first decreased by 67 is equal to three times the second

Answers

Answer:

The numbers are 10 and 11.

Step-by-step explanation:

Let x and x+1 be consecutive positive integers.

When the problem says that the square of the first decreased by 67 this means x^2-67 and this is equal to three times the second 3(x+1).

So x^2-67=3(x+1) will be our equation.

Next, we solve the equation:

x^2-67=3x+3\n\nx^2-70=3x\n\nx^2-3x-70=0

Solve by factoring

x^2-3x-70= \left(x^2+7x\right)+\left(-10x-70\right)\n\nx^2-3x-70=x\left(x+7\right)-10\left(x+7\right)\n\nx^2-3x-70=\left(x+7\right)\left(x-10\right)=0

Using the Zero Factor Principle: If ab = 0, then either a = 0 or b = 0, or both a and b are 0.

x+7=0\nx=-7

And

x-10=0\nx=10

Because the numbers need to be positive integers, we only take x = 10 as a valid solution.

The numbers are 10 and 11.

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second). (a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

Answers

Answer:

The probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day is P(Y>190)=\frac{1}{e^{(19)/(10)}}\approx 0.1496

Step-by-step explanation:

Let Y be the water demand in the early afternoon.

If the random variable Y has density function f (y) and a < b, then the probability that Y falls in the interval [a, b] is

P(a\leq Y \leq b)=\int\limits^a_b {f(y)} \, dy

A random variable Y is said to have an exponential distribution with parameter \beta > 0 if and only if the density function of Y is

f(y)=\left \{ {{(1)/(\beta)e^{-(y)/(\beta) }, \quad{0\:\leq \:y \:\leq \:\infty}   } \atop {0}, \quad elsewhere} \right.

If Y is an exponential random variable with parameter β, then

mean = β

To find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day, you must:

We are given the mean = β = 100 cubic feet per second

P(Y>190)=\int\limits^(\infty)_(190) {(1)/(100)e^(-y/100) } \, dy

Compute the indefinite integral \int (1)/(100)e^{-(y)/(100)}dy

(1)/(100)\cdot \int \:e^{-(y)/(100)}dy\n\n\mathrm{Apply\:u \:substitution}\:u=-(y)/(100)\n\n(1)/(100)\cdot \int \:-100e^udu\n\n(1)/(100)\left(-100\cdot \int \:e^udu\right)\n\n(1)/(100)\left(-100e^u\right)\n\n\mathrm{Substitute\:back}\:u=-(y)/(100)\n\n(1)/(100)\left(-100e^{-(y)/(100)}\right)\n\n-e^{-(y)/(100)}

Compute the boundaries

\int _(190)^(\infty \:)(1)/(100)e^{-(y)/(100)}dy=0-\left(-\frac{1}{e^{(19)/(10)}}\right)

\int _(190)^(\infty \:)(1)/(100)e^{-(y)/(100)}dy=\frac{1}{e^{(19)/(10)}}\approx 0.1496

The probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day is P(Y>190)=\frac{1}{e^{(19)/(10)}}\approx 0.1496
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