Let f(x)equals=33xminus−1, h(x)equals=startfraction 7 over x plus 5 endfraction 7 x+5 . find (hcircle◦f)(66).

1 answer:

7 0

$\bf \begin{cases}
f(x)=3x-1\\
h(x)=\cfrac{7}{x+5}\\\\
(h\circ f)(x)=h(~~f(x)~~)
\end{cases}
\\\\\\
f(66)=3(66)-1\implies \boxed{f(66)=197}
\\\\\\
\stackrel{h(~~f(x)~~)}{h(~~f(66)~~)}\implies \stackrel{h(~~f(66)~~)}{h\left(~~\boxed{197}~~ \right)}=\cfrac{7}{\boxed{197}+5}\implies h\left(~~\boxed{197}~~ \right)=\cfrac{7}{202}$

You might be interested in

What is the equation of the graph below? A graph shows a parabola that opens up and crosses the x axis at negative two and negat

Anastaziya [24]

Answer:

$y=(x+3)^{2} -1$

Step-by-step explanation:

General equation of a parabola that opens up is $y=a(x-h)^{2} +k$ , a>0

So equation becomes $y=(x-h)^{2} +k$ which implies option A and option B can never be true for this parabola.

Also this parabola's vertex lies in 3rd quadrant where coordinates of both x and y will be negative.

So, the equation of parabola will be of the form $y=(x+h)^{2} - k$

Hence, option 4 that is $y=(x+3)^{2} -1$ is correct.

4 0

2 years ago

Read 2 more answers

Maria throws a softball straight up into the air with an initial velocity of 50 ft/sec. The ball leaves her hand when it is exac

Nataly_w [17]

Answer:

The equation that models the situation is $s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$ .

Step-by-step explanation:

Let suppose that effects from air friction and Earth's rotation can be neglected, so that the softball can be modelled experiment a free fall, that is, an uniform accelerated motion due to gravity. From we know the initial velocity and position of the position and we can determine the final position of the ball as a function of the final velocity:

$v_{f}^{2} = v_{o}^{2}+2\cdot a \cdot (s_{f}-s_{o})$ (1)