CPhill

We can find the longer leg - call it L -  with the Law of Sines

L/sine 52 = 140 / sine 90

L  =140 sine(52) = 110.32 m

And the other leg is

140sin(38) = 86.19m

So the perimeter is

140 + 110.32 + 86.19 = 336.51 m

We have

1 = 100e^(-.18t) divide both sides by 100

.01 = e^(-.18t)    take the log of both sides

ln . 01 =ln e^(-.18t)    and since ln e = 1, we can write

ln .01  = (-.18t)       divide both sides by -.18

ln .01 / (-.18)  = t = about 25.58 hrs.

I think this is supposed to be  .... (y+3)/y = 77/56

Cross-multiply

56(y+ 3)  = 77y    simplify

56y + 168  = 77y     subtract 56y from both sides

168 = 21y    divide by 21 on both sides

8 = y

She makes 60 cents selling each cupcake.

So.....the number sold times the profit on each gives the total profit.

And we have....

.60c ≥ $30   where "c" is the number sold

......................................................................................................................

C = (5/9)(F - 32)

Well...there are many problems that have remained unsolved......so any one of them might be the "hardest."  

But....one of the most difficult to solve was a problem known as "Fermat's Last Theorem"...it states that no known unique integers a,b,c exist such that a^n + b^n = c^n for n≥ 3.

This problem took more than 350 years to solve. {At least one hefty sum of money was offered for its solution). The solution was finally found by an English mathematician, Andrew Wiles, in the mid-1990s. Oddly enough, he found it so frustrating that he almost gave up.....!!!

BTW.....Wiles' proof is over 150 pages in length.....!!! Definitely not "light reading"

Even if we added the 1 and 2, first, we still have

6/2(3)  ...but the order of operations dictates that we perform multiplications and divisions at the same level of precedence as we encounter them from left to right....

And  since we encounter the division first, we have 6/2 = 3

And then

3(3) = 9

My advice??? ...Don't rely too much on those "intelligent friends and associates" for a lot of heavy-duty math guidance .... ( LOL..!!!)

This isn't possible.......both numbers have to be positive......and the largest possible result of multiplying two positive numbers that sum to 14 is 49.......that is, when x and y both = 7.

We can first count sets and then arrangements of those sets

We want to choose 1 of the 47 hip-hop MP3s and any 5 of the 14 Celtic Folk MP3s

The total number of sets of songs is given by C(47,1)* C(14,5) = 94094

But......each set of songs can be arranged in 6! = 720 ways

So   94094 x 720 = 67,747,680 different programs