Hello, Twan!

Beth can walk six miles in the same time she can bike 32 miles.

If her biking rate is one more than five times her walking rate, how fast does she walk?

We will use: .$\displaystyle \text{Distance} \:=\:\text{Rate} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{R}$

Let: .$\displaystyle \begin{array}{ccc} r &=& \text{walking rate} \\ 5r+1 &=& \text{biking rate} \end{array}$

She walks 6 miles at $\displaystyle r$ mph.

. . This takes her: .$\displaystyle \frac{6}{r}$ hours.

She bikes 32 miles at $\displaystyle 5r+1$ mph.

. . This takes her: .$\displaystyle \frac{32}{5r+1}$ hours.

These times are equal: . $\displaystyle \frac{6}{r} \:=\:\frac{32}{5r+1}$

Now solve for $\displaystyle r.$

Zack is standing near an 8-foot street lamp.

His distance from the base of the lamppost is two feet more than his height.

The length of his shadow is four times his height.

How tall is Zack? Code:

C *
| *
| * A
| o
8 | | *
| h| *
| | *
* - - - - - o - - - - - - - *
D h+2 B 4h E

Zack is $\displaystyle AB = h.$

The street lamp is $\displaystyle CD = 8.$

We are told that: .$\displaystyle DB \:=\:h+2,\;BE \:=\:4h.$

We see that: $\displaystyle DE \:=\:5h+2$

From the similar right triangles: .$\displaystyle \frac{DE}{CD} \,=\,\frac{BE}{AB} \quad\Rightarrow\quad \frac{5h+2}{8} \:=\:\frac{4h}{h} $

We have: .$\displaystyle \frac{5h+2}{8} \:=\:4 \quad\Rightarrow\quad 5h+2 \:=\:32 \quad\Rightarrow\quad 6h \:=\:30$

Therefore: .$\displaystyle h \:=\:6$ . . . Zack is 6 feet tall.