Trigonometry Help for Hw. Due Tomorrow....

1) f(x)+x^2 and g(x)=2^x are drawn where the unit of measurement is 1 inch. Show that at a distance 2 feet to the right of the orgin, the height of the graph of F is 48ft and the height of g is 265mi...
*does this mean i just convert things to inches and plug it in for x?*

2)a) e^(2ln5) b)ln(lne^e^10))

3) e^(2x) - 3e^(x) +2 = 0

4) a)ln(lnx)=1 b)e^(ax)=Ce^(bx) where a doesnt = b

5) a)sin^-1(sqrt3/2) b)cos^-1 (-1)

6) a)tan^-1(1/sqrt3) b)sec^-1(2)

7) a)arctan(1) b)sin^-1(1/sqrt2)

8) a)cot^-1(/sqrt3) b)arccos(-1/2)

9) a)tan(arctan10) b)sin^-1(sin(7pi/3))

10) a)tan(sec^-1(4)) b)sin(2sin^-1(3/5))

thank you for your help! im kinda confused but have a rought idea of how to tackle these problems....

*BEST ANSWER WILL BE GIVEN TO THOSE WHO HELP*

You asked a lot of questions. I suggest in the future, you limit how many questions you ask at one time. I know it will cost you more points, but think about the responder who spends a lot of time answering just 1 question for a maximum of 12 potential points, when they could answer 12 questions for a maximum of 144 points. Also, what happens if you decide to delete the question while we're answering your question, or after we answer your question?

Now that I'm done preaching:

1)
f(x) = x + 2
g(x) = 2^x

At a distance of 2 feet to the right of the origin, we have moved 24 units (1 unit = 1 inch and 2 feet = 24 inches), so we're looking for f(24) and g(24)

f(24) = 24 + 2 = 26. That should be 26 inches, or 2 feet 2 inches
g(24) = 2^24 = 16777216 inches
16777216 inches =>
1398101.3333333333333333333333333 feet =>
264.79191919191919191919191919192 miles

2)
e^(2 * ln(5)) =>
(e^(ln(5))^2 =>
5^2 =>
25

ln(ln(e^e^(10))) =>
ln(e^(10) * ln(e)) =>
ln(e^(10) * 1) =>
ln(e^(10)) =>