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marysya [2.9K]
3 years ago
14

Let f(x)=x^2 +2 and g(x)=1-3x. find each function value: (fg)(-1)

Mathematics
2 answers:
sammy [17]3 years ago
6 0

Answer:

12

Step-by-step explanation:

so basically

(x^2+2)*(1-3x)=-3x^3+x^2-6x+2

now plug in -1

-3(-1)^3+(-1)^2-6(-1)+2=12

hope this helped :)

photoshop1234 [79]3 years ago
3 0

The correct value of (fg)(-1) is "12". A further solution of the given query is provided below.

Given functions are:

f(x) = x^2+2

g(x) = 1-3x

Now,

⇒ (fg) = (x^2+2)(1-3x)

By applying multiplication, we get

           =x^2+2-3x^3-6x

           =-3x^3+x^2-6x+2...(equation 1)

By substituting the value "-1" in place of "x" in equation 1, we get

⇒ (fg)(-1) = -3(-1)^3+(-1)^2-(6)(-1)+2

                  =-3(-1)+1+6+2

                  =3+1+6+2

                  =12

Thus the right answer is (fg)(-1) = 12.

Learn more:

brainly.com/question/10057660

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suppose that incomes of families in Newport Harbor are normally distributed with a mean of $750,000 and a standard deviation of
denis23 [38]

Answer: 0.345

Step-by-step explanation:

Given : The incomes of families in Newport Harbor are normally distributed with Mean : \mu=\$ 750,000 and Standard deviation : \sigma= $250,000

Samples size : n=4

Let x be the random variable that represents the incomes of families in Newport Harbor.

The z-statistic :-

z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For x= $800,000

z=\dfrac{800000-750000}{\dfrac{250000}{\sqrt{4}}}=0.4

By using the standard normal distribution table , we have

The probability that the average income of these 4 families exceeds $800,000 :-

P(x>80,000)=P(z>0.4)=1-P(z\leq0.4)\\\\=1-0.6554217=0.3445783\approx0.345

Hence, the probability that the average income of these 4 families exceeds $800,000 =0.345

6 0
4 years ago
Solve using elimination<br> x+y-2z=8<br> 5x-3y+z=-6<br> -2x-y+4z=-13
Free_Kalibri [48]
So here is your answer with LaTeX issued format interpretation. Full process elucidated briefly, below:

\begin{alignedat}{3}x + y - 2z = 8 \\ 5x - 3y + 2 = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}

For this equation to get obtained under the impression of those variables we have to eliminate them individually for moving further and simplifying the linear equation with three variables along the axis.

Multiply the equation of x + y - 2z = 8 by a number with a value of 5; Here this becomes; 5x + 5y - 10z = 40; So:

\begin{alignedat}{3}5x + 5y - 10z = 40 \\ 5x - 3y + z = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}

Pair up the equations in a way to eliminate the provided variable on our side, that is; "x":

5x - 3y + z = - 6

-

5x + 5y - 10z = 40
______________

- 8y + 11z = - 46

Therefore, we are getting.

\begin{alignedat}{3}5x + 5y - 10z = 40 \\ - 8y + 11z = - 46 \\ - 2x - y + 4z = - 13 \end{alignedat}

Multiply the equation of 5x + 5y - 10z = - 40 by a number with a value of 2; Here this becomes; 10x + 10y - 20z = 80.

Multiply the equation of - 2x - y + 4z = - 13 by a number with a value of 5; Here this becomes; - 10x - 5y + 20z = - 65; So:

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ - 10x - 5y + 20z = - 65 \end{alignedat}

Pair up the equations in a way to eliminate the provided variables on our side, that is; "x" and "z":

- 10x - 5y + 20z = - 65

+
10x + 10y - 20z = 80
__________________

5y = 15

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ 5y = 15 \end{alignedat}

Multiply the equation of - 8y + 11z = - 46 by a number with a value of 5; Here this becomes; - 40y + 55z = - 230.

Multiply the equation of 5y = 15 by a number with a value of 8; Here this becomes; 40y = 120; So:

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 690 \\ 40y = 120 \end{alignedat}

Pair up the equations in a way to eliminate the provided variables on our side, that is; "y":

40y = 120

+

- 40y + 55z = - 230
_________________

55z = - 110

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 230 \\ 55z = - 110 \end{alignedat}

Solving for the variable of 'z':

\mathsf{55z = - 110}

\bf{\dfrac{55z}{55} = \dfrac{-110}{55}}

Cancel out the common factor acquired on the numerator and denominator, that is, "55":

z = - \dfrac{\overbrace{\sout{110}}^{2}}{\underbrace{\sout{55}}_{1}}

\boxed{\mathbf{z = - 2}}

Solving for variable "y":

\mathbf{\therefore \quad - 40y - 55 \big(- 2 \big) = - 230}

\mathbf{- 40y - 55 \times 2 = - 230}

\mathbf{- 40y - 110 = - 230}

\mathbf{- 40y - 110 + 110 = - 230 + 110}

Adding the numbered value as 110 into this equation (in previous step).

\mathbf{- 40y = - 120}

Divide by - 40.

\mathbf{\dfrac{- 40y}{- 40} = \dfrac{- 120}{- 40}}

\mathbf{y = \dfrac{- 120}{- 40}}

\boxed{\mathbf{y = 3}}

Solve for variable "x":

\mathbf{10x + 10y - 20z = 80}

\mathbf{Since, \: z = - 2; \quad y = 3}

\mathbf{10x + 10 \times 3 - 20 \times (- 2) = 80}

\mathbf{10x + 10 \times 3 + 20 \times 2 = 80}

\mathbf{10x + 30 + 20 \times 2 = 80}

\mathbf{10x + 30 + 40 = 80}

\mathbf{10x + 70 = 80}

\mathbf{10x + 70 - 70 = 80 - 70}

\mathbf{10x = 10}

Divide by this numbered value \mathbf{10} to get the final value for the variable "x".

\mathbf{\dfrac{10x}{10} = \dfrac{10}{10}}

The numbered values in the numerator and the denominator are the same, on both the sides. This will mean the "x" variable will be left on the left hand side and numbered values "10" will give a product of "1" after the division is done. On the right hand side the numbered values get divided to obtain the final solution for final system of equation for variable "x" as "1".

\boxed{\mathbf{x = 1}}

Final solutions for the respective variables in the form of " (x, y, z) " is:

\boxed{\mathbf{\underline{\Bigg(1, \: \: 3, \: \: - 2 \Bigg)}}}

Hope it helps.
8 0
3 years ago
Read 2 more answers
The product of -5 and the sum of X and eight decreased by the product of three and X
Alla [95]
-5(x+8)-3x

simplified would be
-8x-40
5 0
3 years ago
Write the sentence as an equation.<br> 143 more than the product of g and 369 is equal to 141.
DiKsa [7]

Answer: 143>(369g=141)

Step-by-step explanation:

143>(369g=141)

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3 years ago
Robert is building model cars as well as model trains. The cars take 4 hours to build while the trains take 6 hours
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10 cars and 10 trains in 100 hours

it approximately take him 10 hours to build one car and one train how 6+4=10 how many times can 10 go into 100 10 times hope this helps a little

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