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Definitely easier!

Not really... mind the leap year (Like EP said, 2004 is a leap year )

Yeah, sorry I didn't see this one before I posted. Go with this one. More likely to be correct.

Sorry, I don't get the problem that well. Here is my best shot at it:

There are 18 years and 12 months in each year. In order to calculate the amount warned per month, we will have to compute: \((75000/18)/12 \implies 9875/27\)

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After doing this, set x as the amount that has to be paid: \(.039x=9875/27\)

Then just compute:\(x=9377.9677113\)

Note: Electric Pavlov's answer is far more likely to be correct, so if ours are different, go with his/her's

Assuming this is an annuity due (payments at beginning of month)

FV = P [(1+i)ⁿ -1]/i  *  (1+i)      i = .039/12     n = 18 yr * 12 months/yr = 216

P = 252.01     

Let Δ ABC be the right triangle with AB = 18 and BC = 16.         18 * 16 / 2 = 144 u²

∠ADB = arctan(AB / BD) 77.471º

∠GDC = 180 - 77.471 = 102.529º

∠BCF = arctan(BF / BC) = 41.987º

3 more steps:

1/   Use the law of sines to calculate the length of a side GC

2/   Use trigonometry to calculate the height from D to GC

3/   Multiply side GC by the height and then divide by 2

Have fun!       (btw, the answer is not 108) 

.10 (x)    +  1.00 (39-x)  = .40  (39)

-.9x = -23.4

x = 26 liters of 10%    13 liters of pure

(a) \(\frac{d}{dx}7x^{-1}=\boxed{-7x^{-2}}\) by the power rule

I don't understand the other problems 

Let there be a liters of full acid, b liters of 10% acid. We have \(a+b=39, \frac{a+0.1b}{a+b}=0.4\rightarrow a+0.1b=15.6\rightarrow \boxed{a=13, b=26}\)

\(\left(\frac{2}{a}+\frac{1}{b}\right)^2=\frac{4}{a^2}+\frac{1}{b^2}+\frac{4}{ab}=81\rightarrow 57+\frac{4}{ab}=81\rightarrow \frac{4}{ab}=24\rightarrow \frac{1}{6}=ab\rightarrow\boxed{3ab=\frac{1}{2}}\)

do you mean 222?

This is a typical AMC 10/12 problem. We apply Mass points. Let C have mass 4. B will then have mass 4*3=12. A will then have mass 12*4=48. So F will have mass 12+48=60. So FG:GC=1:15. The height of GDC is 15/16 the large triangle, and the base is 4/5 the large triangle. Our answer is 144*15/16*4/5=108

How do you necessarily know he was cheating hm?

The answer is 365+365+365+366=1461 days.

365 + 365 + 365 + 366                  ( 2004 was a leap year)

1461 days

(x - (sqrt3 /2 + 1/2) ( x - (sqrt3 /2 -1/2 )        expand via distributive property and simplify

x²  - sqrt3/2 x + 1/2 x         - sqrt3 /2 x  + 3/4  - sqrt3/4       - 1/2x + sqrt3 /4  -1/4

x^2 - sqrt3 x + 1/2   

The answer is C.

There were a total of 1462 days.

25 thru 250 = 226 numbers

perfect cubes   27  64  125  216           222 are not perfect cubes         *** edited***   (oops!)

Daunting.

\(\frac{\frac{a^2}{x-a}+\frac{b^2}{x-b}+\frac{c^2}{x-c}+a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}=\frac{a+b+c}{\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}}+a+b+c=\frac{a+b+c}{\frac{a(x-b)(x-c)+b(x-a)(x-c)+c(x-a)(x-b)}{(x-a)(x-b)(x-c)}}+a+b+c\)

I made this answer, before creating my account. How do I claim this answer?

Much easier:

Note that \(d=\frac{c_2-c_1}{\sqrt{A^2+B^2}}\) for lines \(ax+by+c_1=0, ax+by+c_2=0\). So we have a=-1/2, b=1, c1=-1, c2=4. Plugging, we get \(\frac{4+1}{\sqrt{1/4+1}}=\frac{5}{\sqrt{5/4}}=\frac{5}{\frac{\sqrt{5}}{2}}=\frac{10}{\sqrt{5}}=\frac{10\sqrt{5}}{5}=\boxed{2\sqrt{5}}\) . 

Note: This yields the same answer as @asinus got, just in exact form

Note that \(46656=2^6*3^6\). We take the square root of both sides to get xy=2^3*3^3. So we are essentially finding the sum of the factors of 2^3*3^3. S(2^3*3^3)=S(2^3)*S(3^3). For 2^3 --> 1,2,4,8 --> 15. For 3^3 --> 1,3,9,27 --> 40. So 15*40=600, which is your answer.

Let n be the number of nickels and d be the number of dimes. We have

n+d=54 --> 5n+5d=270

5n+10d=425

5d=425-270=155 --> d=31 --> n=23