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$c=1/2e$(c is calvin's money at the start, e is eddy's money at the start)

$8(c-36)=e$

Substituting $c=1/2e$

$8(1/2e-36)=e$

$4e-288=e$

$3e=288$

$e=96$ dollars

$c=1/2e$

$c=1/2(96)$

$c=48$ dollars

$48+96=\boxed{144}$ dollars

We have:

$c=\frac12 e$, $8(c-36)=e$. Now solve for $e+c$.

The point is something like $(x,\frac{1}{2}x^2-9)$ for some constant value of $x$. The distance between this and the origin is $a=\sqrt{x^2+(\frac{1}{2}x^2-9)^2}$, and so $a^2=x^2+(\frac{1}{2}x^2-9)^2=x^2+\frac{1}{4}x^4-9x^2+81=\frac{1}{4}x^4-8x^2+81=\frac14(x^4-32x^2+324)$ and factor the inside as $(x^2+a)^2+b$ which has minimum value of $b$. Then multiply by $\frac14$ and finish. :)

The number 3456 does add up to 18 so it would be 3456

3465

a_n = 5 + (n-1)   Arithnetic Sequence

   sum₁₀ = n/2 ( a₁+a₁₀)= 10/2 ( 5 + 14) = 190/2 = 95 coins saved in 10 weeks

It's equal to $(3^{33}+7^{54})(8^{326}+(-1)^{2016})\pmod{10}$.

$3,7,8$ all cycle modulo $10$ and $(-1)^{2016}$ is trivial to deal with.

If I understand the question correctly our equation is (5*0.5) + (6*0.5) + (7*0.5) + (8*0.5) + (9*0.5) + (10*0.5) + (11*0.5) + (12*0.5) + (13*0.5) + (14*0.5) = $47.5.

He saves 95 fifty-cent coins in 10 days           

Thank you but the answer was 1337. 

Thank you, but it's incorrect. 

= \(\frac{1}{3}\times \pi \times (16-12)\)

= \(\frac{4}{3}\pi\)

Here ya go :

What is the average of all positive integers that have four digits when written in base 3, but two digits when written in base 6? Write your answer in base 10.

4 digits base 3

smallest 1000 = 3^3 = 27

biggest  2222 = 2*27+2*9+2*3+2 = 80

3 digits base 6

smallest 100 = 36

biggest  1000-1 = 6^3-1 = 4095

So the intersection of these is    36 to 80  there are  80-36+1 = 45 of them

sum of numbers from 36 to 80 = (45/2)(72+44) = 45/2*116= 45*58 = 2610

average = 2610/45 =58

A more straight forward approach would be to say that of there are 45 numbers the median on is in the 46/2 = 23rd position.  In this case the median and the mean will be the same.

36+23-1 = 58   Just as expected.

So the answer is 58 base 10

Ok that seems good

1 orange + $0.10

(1 orange + $0.10) * 15
= 15 pears + ( 15 * $0.10)
= 15 pears + $1.50

15 pears + $1.50 = 18 oranges (Cost of 15 pears = Cost of 18 oranges)

18 oranges - 15 pears = 3 oranges

3 oranges —> $1.50
1 orange —> $0.50 
1 pear —> $0.50 + $0.10 = $0.60
15 pears —> $0.60 * 15 = $9
Willy spent $9 on the 15 pears.

15 (r+10) = 18 (r)      r = price of orange

150 = 3r

r = 50       so 18 oranges is   9 dollars   ( the same price as 15 pears)

HINT: Write cost of orange as cost of pear - $0.10$ and set up a system of equations.

By multiplying the minutes by 3, so you go three times the distance, so $25*3=75$ km

f = 5 bills                          value =  5f

t = 10 bills = 20-f              value =  10(20-f)                    summed = 125

5f         +       10(20-f)         = 125

-5f  = -75          You can take it from here .....

The largest number will be the digit that starts with the largest number and goes down the chain, which would be $96521$.

But wait! They have to be even! So the largest number would be $96512$. The next largest number would use 2 as the hundreds digit, so it would be $96251$.

But wait! It has to be even again! Because we used up the 6 and the two already, we need to have 1 in the hundreds place: $96152$.

So now, our list of numbers is $96512,96152$. So the third even number would use 5 as a thousands digit, so $95621$, but that is not even! It would be $\boxed{95612}$

ohhh oops, my bad. 

Yea, you're right. :))

=^._.^=

How you get 75 Km?

If it's $25$ km in $20$ minutes, then it is $75$ km in $60$ minutes which is $1$ hour, so Jason drove at $75$ kmph, which is above the speed limit.