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3 years ago

Let f(x) = 2x – 1, g(x) = 3x, and h(x) = x2 + 1. Compute the following.

1 answer:

4 0

4.
h(f(x)=h(2x-1)=
(2x-1)^2+1=
4x²-4x+1+1=
4x²-4x+2

5.
f(f(x))=2(2x-1)-1=4x-2-1=4x-3

6.
f o g (x)=f(g(x))
h o g (x)=h(g(x))=
(3x)^2+1=
9x²+1

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(5,12) with a slope m=10 the equation of the line is

motikmotik

Answer:

y= 10x-38

Step-by-step explanation:

Equation of a line is usually written in the form of y=mx+c, where m is its gradient and c is its y-intercept.

Given that m=10,

y= 10x +c

Now substitute the coordinates into the equation.

When y=12, x=5,

12= 10(5) +c

12= 50 +c

c= 12 -50

c= -38

Thus the equation of the line is y= 10x -38

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3 years ago

If each month in reno had the same average rainfall as in august, what would the total number of millimeters be after 12 months

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Answer:

12 times the August rainfall

Step-by-step explanation:

The total of 12 of the same value is 12 times that value. Multiply the August average by 12 to get your answer.

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3 years ago

What polynomial must be added to 3x2 + 4x + 7 to obtain the sum of 0?

Dovator [93]

Answer:

do (3x2+4x+7)+y=0

y=-3x^2-4x-7

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3 years ago

A boat traveled 210 miles downstream and back. The trip downstream took 10 hours. The trip back took 70 hours. What is the speed

musickatia [10]

Try this solution:

Answer: 6 & 4.5 (m-per-h)

Step-by-step explanation:

- The basic formula used in this task is S=V*t, where S - distance, V - speed, t - time.

- the downstream speed is: V+Vc, where V - the speed of the boat in still water, Vc - the speed of the current.

- the upstream speed is: V-Vc.

- according to the described above the distance for downstream is S1=(V+Vc)*t1, where t1=10; the distance for upstream is S2=(V-Vc)*t2, where t2=70.

- for whole travel down- and upstream: S=S1+S2.

- Using these it is possible to make up the system of the equations:

$\left \{ {{(70(V-V_c)=10(V+V_c)} \atop {70(V-V_c)+10(V+V_c)=210}} \right.$

V - the speed of the boat in still water - 6 miles per hour, Vc - the speed of the current - 4.5 miles per hour. All the details for the system of the equations are in the attachment.

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3 years ago

For the cost function c equals 0.1 q squared plus 2.1 q plus 8, how fast does c change with respect to q when q equals 11? Det

Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is 9.95%

Step-by-step explanation:

Cost function is given as, $c=0.1\:q^{2}+2.1\:q+8$

Given that c changes with respect to q that is, $\dfrac{dc}{dq}$ . So differentiating given function,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)$

Applying sum rule of derivative,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)$

Applying power rule and constant rule of derivative,

$\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0$

$\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1$

$\dfrac{dc}{dt}=0.2\left(q\right)+2.1$

Substituting the value of $q=11$ ,

$\dfrac{dc}{dt}=0.2\left(11\right)+21.$

$\dfrac{dc}{dt}=2.2+2.1$

$\dfrac{dc}{dt}=4.3$

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,

$Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100$

Rewriting in terms of cost C,

$Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100$

Calculating value of $C\left(q \right)$

$C\left(q\right)=0.1\:q^{2}+2.1\:q+8$

Substituting the value of $q=11$ ,

$C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8$

$C\left(q\right)=0.1\left(121\right)+23.1+8$

$C\left(q\right)=12.1+23.1+8$

$C\left(q\right)=43.2$

Now using the formula for percentage,

$Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100$

$Percentage\:rate\:of\:change=0.0995\times 100$

$Percentage\:rate\:of\:change=9.95%$

Percentage rate of change of c with respect to q is 9.95%

7 0

3 years ago