I totally agree with you, Melody!
As I've been here for about 2 years now! (Wow!), I've seen homework questions been posted multiple times by users, but haven't seen the users type anything back after another user posts. They just read their answer, discarding their solution.
2. We think of a Pythagorean triplet, here:
a^2+b^2=c^2
So, 5^2+12^2=13^2
So, the answer is \(\boxed{2}\)
The current helps the frog swim downstream and hinders her swimming upstream. Let the frog's rate in still water be \(x\) and the current be \(y.\) Thus, the swimming downstream the frog's rate is \(x+y\) , while swimming upstream it is \(x-y\) . Now apply rate times equals distance: downstream, we have
\((x+y)(2)=8\)
and upstream gives
\((x-y)(14)=8\)
Solving these equations, we find that \((x,y)=(16/7, 12/7).\)
Thus the frog's rate in still water is \(\boxed{\frac{16}{7}}\)miles per hour.
knight=1/15
kinght+partner=1/10
partner=1/10-1/15=1/30
partner=30 days
_ _ _ _ _
We have the numbers 2,2,2,9,9
Possibilities with 2: 2,2,2,9,9
2,2,9,2,9, 2,2,9,9,2, 2,9,2,2,9, 2,9,2,9,2, 2,9,9,2,2
We have only four possibilties, because we have only two 9's.
Thus, the answer is \(6+4=\boxed{10}\)
You can also use elimination:
Movies=m, Games=g
So, 2m+3g=24.30
3m+1g=18.25
Find the LCM of 3 and 2, which is 6.
So, 6m+9g=72.90
6m+2g=36.5
Canceling both 6m's, we have:
7g=36.4
g=\(\boxed{5.2}\)
m=4.35
But, since we're asking for the rent of one game, the answer is \(\boxed{5.2}\)
106P2=106!/104!=106*105=11130
Or, is it 106C2=106!/2!*104!=106*105/2*1=5565?
Which one?
0.68=68/100
17/25=68/100
So, no, 0.68 is not greater than 17/25
B sounds good; I think that's the answer.
Solution:
6 foot man=20 feet
6/20=1 foot
?= 50 feet
Multiply 6/20*50=3*5=
\(\boxed{15}\)
