# Thread: [SOLVED] last 3 homework problems im stumpped on

1. ## [SOLVED] last 3 homework problems im stumpped on

1. They want me to evaluate the expression using the values given in table for (fog)(4)

x | 1,5,11,12
f(x)|-4, 11, 3, 15

x |-5, -4, 1, 4
g(x)|1, -7, 5, 11

2.Solve the equation
4^7-3x=1/16 (I know this has something to do with logs, but i dont know what)

3.While traveling in a car, the centrifugal force a passenger experiences as the car drives in a circle varies jointly as the mass of the passenger and the square of the speed of the car. If the passenger experiences a force of 352.8 newtons when the car is moving at a speed of 70 kph and the passenger has a mass of 80 kg, find the force a passenger experiences when the car is moving at 20kph and the passenger has a mass of 90kg.

Im not asking for the answers, i just dont know how to solve the problems. The formulas, steps, or functions are what I am looking for.

2. Although I'm not too sure on what the last two problems are asking for, I can help with the first problem.

When you see something like $\displaystyle (f\circ g)(4)$, what essentially is being asked is the evaluation of $\displaystyle f(g(4))$. So find the evaluation of $\displaystyle g(4)$ and then find the evaluation of $\displaystyle f(x)$ at the value $\displaystyle x=g(4)$.

3. Originally Posted by Pinkk
Although I'm not too sure on what the last two problems are asking for, I can help with the first problem.

When you see something like $\displaystyle (f*g)(4)$, what essentially is being asked is the evaluation of $\displaystyle f(g(4))$. So find the evaluation of $\displaystyle g(4)$ and then find the evaluation of $\displaystyle f(x)$ at the value $\displaystyle x=g(4)$.
i dont think im following. How do you evaluate it from a table? What im thinking is that in the table of g(x), when g is 11, x=4
Is that what you're refering to?

4. When looking at the chart, first look at what $\displaystyle g(x)$ is when $\displaystyle x=4$. In the chart, $\displaystyle g(4)=11$. Now you look at what $\displaystyle f(x)$ is when $\displaystyle x=11$, and in the chart, $\displaystyle f(11)=3$. Notice how $\displaystyle g(4)=11$, so what you have is $\displaystyle (f\circ g)(4)=f(g(4))=f(11)=3$.

5. Hello, ninobrn99!

Using the values in the tables, evaluate: .$\displaystyle (f\circ g)(4)$

. . $\displaystyle \begin{array}{c|cccc} x & 1 &5 &11 & 12 \\ \hline f(x)& \text{-}4 & 11 & 3 & 15\end{array} \qquad\qquad \begin{array}{c|cccc}x & \text{-}5&\text{-}4 & 1 & 4 \\ \hline g(x)& 1 & \text{-}7 & 5 & 11 \end{array}$
$\displaystyle (f\circ g)(4)$ means: .$\displaystyle f(g(4))$

From the second table: .$\displaystyle g(4) = 11$

. . So: .$\displaystyle f(g(4)) \;=\;f(11)$

From the first table: .$\displaystyle f(11) = 3$

. . Therefore: .$\displaystyle f(g(4)) \:=\:3$

2. Solve the equation: .$\displaystyle 4^{7-3x}\:=\:\frac{1}{16}$

We have: .$\displaystyle 4^{7-3x} \:=\:\frac{1}{4^2} \quad\Rightarrow\quad 4^{7-3x} \:=\:4^{-2}$

Hence: .$\displaystyle 7-3x \:=\:-2 \quad\Rightarrow\quad -3x \:=\:-9 \quad\Rightarrow\quad x \:=\:3$

6. Okay, thank you both. After seeing it, I was on the right path, i just couldn't figure out where I was hanging up. Thank you...now just the troublesome word problem

7. No problem. I suggest getting used to the [ math ] tagging with problems, because I thought your second problem was $\displaystyle 4^{7} - 3x = \frac{1}{16}$, not $\displaystyle 4^{7-3x}=\frac{1}{16}$

8. Originally Posted by Pinkk
No problem. I suggest getting used to the [ math ] tagging with problems, because I thought your second problem was $\displaystyle 4^{7} - 3x = \frac{1}{16}$, not $\displaystyle 4^{7-3x}=\frac{1}{16}$
sorry, i thought the problem was
the same as you because the -3x didnt look like a superscript. My bust on that and yes, i am practicing on my Latex. Thanks again.

9. okay, i think i figured out the word problem

okay...so the formula is
$\displaystyle F=\frac{mv^2}{r}$
with that, I need to figure out what r is in order to figure out the second part of the problem.
$\displaystyle 352.8=\frac{(80)(70)^2}{r}$
so now i try to solve for r:
$\displaystyle 352.8r=(80)(4900)$

$\displaystyle 352.8r=392000$
$\displaystyle r=1111.11$

With r figured out, i can now solve the following:
$\displaystyle F=\frac{(90)(20)^2}{1111.11}$

$\displaystyle F=32.4 newtons$

I know the latex is all messed up, but does that look right?