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Let E be what he earns per month

So

E - 120   - 95  =

E - 215            is what he has left after   the food and  transport

And then he spends  (2/5) of what's left

(2/5)(E -215)

So

(E  -215) - (2/5) (E - 215)  = 780

(3/5)  (E -215)  = 780

E  -  215  =  780 (5/3)

E  - 215  =  1300

E =  1300 + 215  =   $1515

S +  M  =  59  (1)

M + I    =   52 (2)

S + I   =   41   (3)

Subtract  (3)  from (1)

M - I    =   18     (4)      and

M + I  =    52     (2)

Add (4) and (2)

2M =  70

M  = 35

S + M = 59

S + 35 = 59

S = 59 - 35  =  24

M + I   = 52

35 + I  = 52

I  = 52  - 35   =   17

oh yeah my teacher showed me it once. very fun to use, mostly goes over fundamentals

SSA Congruence is not a valid Congruence Theorem.

2a + 2b + 2c = 6

If you write out several of the terms:   3, 5, 2, -3, -5, -2,    3, 5, 2, -3, -5, -2,    3, 5, 2, ...

you can notice that the sum of the first six terms is 0, 

the sum of the next six terms is 0, etc.

Dividing 1616 by 6 gives a quotient of 269 and a remainder of 2.

This means that there are 269 groups each of which sum to 0

and two remaining terms 3 and 5 which sum to 8.

The sum of all thee terms is 8.

For an integer to be relatively prime to 60, it has to be:

(i) not divisible by 2

(ii) not divisible by 3

(iii) not divisible by 5

By inclusion-exclusion principle and using the fact that \(\text{# of integers in }[1, N]\text{ divisible by }k = \left\lfloor\dfrac{N}k\right\rfloor\),

\(\begin{array}{cl} &\text{# of numbers relatively prime to 60}\\ =& 1000000 - \left\lfloor\dfrac{1000000}2\right\rfloor - \left\lfloor\dfrac{1000000}3\right\rfloor - \left\lfloor\dfrac{1000000}5\right\rfloor + \left\lfloor\dfrac{1000000}6\right\rfloor + \left\lfloor\dfrac{1000000}{10}\right\rfloor + \left\lfloor\dfrac{1000000}{15}\right\rfloor-\left\lfloor\dfrac{1000000}{30}\right\rfloor \\=&266666\end{array}\)

I think the final answer is wrong, because:

\(\begin{array}{rcl} \text{length of diagonal} &=& \sqrt{\left(\dfrac{5 - 2\sqrt 3}2\right)^2 + \left(\dfrac{5 + 2\sqrt 3}2\right)^2}\\ &=& \dfrac12 \sqrt{(5 - 2\sqrt 3)^2 + (5 + 2\sqrt 3)^2}\\ &=& \dfrac12 \sqrt{(25 - 20\sqrt 3 + 12) + (25 + 20 \sqrt 3 + 12)}\\ &=& \dfrac{\sqrt{74}}2 \\&\neq& \dfrac{5\sqrt 2}2 \end{array}\)

Here's an alternative solution, that may be easier to understand.

\(\begin{array}{rcl} \tan \angle BAN &=& \dfrac{1}{2}\\ \angle BAN &=& \arctan \dfrac12 \end{array}\)

\(\begin{array}{rcl} \tan \angle AMD &=& \dfrac{2}{\frac23}\\ &=& 3\\ \angle AMD &=& \arctan 3 \end{array}\)

We now use the sine rule:

\(\begin{array}{rcl} \dfrac{MP}{\sin \angle BAN} &=& \dfrac{AM}{\sin\left(180^\circ - \arctan \dfrac12 - \arctan 3\right)}\\ \sqrt5 MP&=& \dfrac{5 \sqrt 2}7\cdot\dfrac23\\ MP &=& \dfrac{2\sqrt{10}}{21} \end{array}\)

Then, 

\(\begin{array}{rcl} \text{area of }\triangle AMP &=& \dfrac12 \cdot \dfrac23 \cdot MP \cdot \sin \angle AMP\\ &=& \dfrac{2\sqrt{10}}{63}\cdot \dfrac{3\sqrt{10}}{10}\\ &=& \dfrac2{21} \end{array}\)

Wait... I made an error, I wrongfully concluded that PF was 1/3 BN. MaxWong's answer is correct, and mine is incorrect

Here I provide a solution with coordinate geometry.

Let D = (0, 0), C = (2, 0), A = (0, 2), B = (2, 2), M = (2/3, 2), N = (2, 1).

The equation of DM is:

\(y = \dfrac{2 - 0}{\dfrac23 - 0}x\\\)

i.e., \(y = 3x\)

The equation of AN is:

\(y - 1 = \dfrac{2 - 1}{0 - 2}(x - 2)\\ -2(y - 1) = x - 2\\ x + 2y-4=0\)

We solve for the coordinates of P.

\(\begin{cases}y = 3x\\x+2y-4=0\end{cases}\\ x + 2(3x) - 4 = 0\\ x= \dfrac47\\ y = 3\left(\dfrac47\right) = \dfrac{12}7\)

Hence P = (4/7, 12/7).

Therefore, 

\(\begin{array}{rcl} \text{area of }\triangle AMP &=& \dfrac12\cdot AM \cdot \left(2 - \dfrac{12}7\right) \\ &=& \dfrac12 \cdot \dfrac23 \cdot \dfrac27\\ &=& \dfrac2{21} \end{array}\)

Since \(x \equiv 3 \pmod{20}\), let x = 20k + 3 for some integer k.

\(20k + 3 \equiv 13\pmod{45}\\ 20k \equiv 10 \pmod{45}\\ 4k \equiv 2 \pmod{9}\\ 4k \equiv 20 \pmod{9}\\ k \equiv 5 \pmod 9\)

Then we can let k = 9n + 5 for some integer n. Substituting this into x = 20k + 3 gives x = 180n + 103.

Therefore, taking (mod 6), we have:

\(x \equiv 180n + 103\pmod 6\\ x \equiv 103 \pmod 6\\ x \equiv 1 \pmod 6\)

It turns out the system has a solution if and only if \(r \equiv 1 \pmod 6\), i.e., \(r = 6p + 1\) for some integer p.

Extend point P so it is perpendicular to AB. Label the intersection point F. 

\([AMP] = {2 \over3} \times PF \div 2\)

\(\triangle APF \) is similar to \(\triangle ABN\). Because of this, \(PF = {1 \over3}\)

Thus, \([AMP] = {2 \over 3} \times {1 \over 3} \div 2 = \color{brown}\boxed{1 \over 9}\)

Gahh... Speed is tricky

Happy Birthday Melody! Enjoy your cake! 

Yep!!!!

The time at 60 mph    d/60

The time at 80 mph    d/80

Time for the whole trip     2d / a          where a = average

d/60 + d/80 = 2d/a

a = 2d / ( d/60 + d/80) = 2 / ( 1/60 + 1/80) = 2 / ( 7/240) = 68.57 mph average

The big square has side length of 5. 

The small square has a side length of \(\sqrt {12} = 2\sqrt3\)

The rectangles are congruent, so their width can be expressed as \(2x+2\sqrt3=5\)

Solving for x, we find \(x = {{5 - 2\sqrt3} \over 2}\)

The length can be expressed as \({2\sqrt3}+{5 -2\sqrt3 \over 2} = {5+2\sqrt3 \over2}\)

Using the Pythagorean Theorem, the length of the diagonal is \(\color{brown}\boxed{5\sqrt2 \over2}\)

i=0;a=1;b=1000000;c=60;cycle:d=a;if(gcd(c,d)==1,goto end, goto next);end:print d,", ",;i++; next: a=a+1;if(a<=b, goto cycle, 0);print"Total No. of coprimes ==", i

OUTPUT==266,666 integers between 1 and 1,000,000 that are relatively prime to 60

Happens to the best of us lol...

Thank you so much Alan!

Because it's the same distance, it is \((60 + 80) \div 2 = \color{brown}\boxed{70} \) mph. 

a + b  =   24   ⇒    b = 24  - a

So

2ab + 10a  = 3b + 257

2a (24 -a) + 10a  = 3(24 -a) + 257

48a  -2a^2 + 10a  = 72 - 3a + 257        rearrange as

2a^2 - 61a +  329    = 0         factors as

(2a  - 47) ( a - 7)    = 0         

The second factor set  = 0   gives an  integer for "a"

a - 7  =  0 

a =7

b = 24  - 7  =17

ab  =  (7)(17)   =   119 

Since no one else has posted an answer, I'll post what I have ...

Using 'a' as the side opposite angle(A), 'b' as the side opposite angle(B), and 'c' as the side opposite angle(C):

Since angle(B) is a right angle:  sin(A) = a/b     cos(A) = c/b     tan(A) = a/c     b² = a² + c²

sin(A)  =  2·cos(A) + tan(A)     --->     a/b  =  2(c/b) + a/c

          multiplying by  bc          --->      ac  =  2c² + ab

Solving for  c  by using the quadratic formula:  ac  =  2c² + ab

                                                         2c² - ac + ab  =  0

                                c  =  [ a +/- sqrt( a² - 8ab ) ] /4

Since the answer is a real number:  a² - 8ab  >= 0

                                                                   a²  >=  8ab

Since  a  must be positive:                         a  >=  8b

which means that  a  must be larger than  b  but  b  is the hypotenuse!

So, either the problem, as stated, is impossible or I messed up!  Help!

Note that he can choose (i) 4 green peppers and 2 red peppers, or (ii) 5 green peppers and 1 red pepper. Then,

\(\begin{array}{rcl} \text{Probability} &=&\dfrac{\displaystyle \binom{5}4\binom{10}2+\binom{5}5\binom{10}1}{\displaystyle \binom{10+5}6}\\ &=& \dfrac{47}{1001} \end{array}\)