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Multiply first equation by 3,

\(3x + 9y = -12\)

\(3x -7y = 36\)

Subtract,

\(16y = -48\)

\(y = -3\)

Plug y back into one of the equations,

\(x -9 = -4 \)

\(x = 5\)

So (x, y) = (5, -3)

I also have had this problem for alcumus- it was a very challenging one ! I eventually solved it, so I can give you a hint:

Consider the possible ways to make a square. Find a pattern in each of every one of them. Add them up to form an equation, and the answer is your answer.

I will post a solution later

Here

when I click on the link the page says "404 ERROR"

http://www.stutzfamily.com/mrstutzblog/wp-content/uploads/2012/02/2009_Chapter_Solution_Set_2009_-_v2.pdf 

Page 7

Solving for $x$ in $x+y=9$ yields $x=9-y$. Then plugging in this value of $x$ in the second expression yields $(9-y)y=18$. Solving for $y$ yields $y=3$ or $y=6$. Then we have the solutions $x=6$ and $x=3$ respecitvely. Now both sums are the same, so we have $x^3+y^3=6^3+3^3=243$.

In one method, you can use x and y, and figure out based on the expansion of (x+y)^3

But you can also notice that an x and y that satisfy \(x+y=9\) and \(xy=18\) is (6, 3)

Plugging these values in,

\(x^3+y^3=216+27=243\)

\(\)

The 2 two boys  can  be arranged in  2 ways   and they can fill any of  4 spots 

And  for each of these, the girls can be arranged in 3! = 6  ways

So

2 * 4 * 6   =    48  ways

Oh I completely forgot about arranging the two boys, thank you!

X^2   - x  - 2   = (x -2)  ( x + 1)

Multiply through by   ( x -2) ( x + 1)    and we have  that

2x + 17 =   a( x + 1)  +  b ( x + 2)

2x + 17  =  (a + b)x  + (a   + 2b)         which implies that

a +  b = 2       ⇒    - a   - b   =  -2        (1)

a + 2b = 17          (2)

Add (1)  and (2)

b  =  15

a  + b = 2   ⇒    a + 15  = 2  ⇒   a  =   2 -15   ⇒  a  =   -13

There are two pints per quart

3 q *  2 p/qt  = 6 pints needed

6 pint /1.5 pint/container = 4 containers

thanks heureka!

The two boys can stand next to each other in 8 ways in the line of 5

  then fill the other three slots with the girls   3 x 2 x 1

 8 x 3 x 2 x 1 = 48 ways

If the 2 boys must stand next to each other, you can treat them as one thing. That means there are 4 things to arrange, which is just 4! = 24

THX,  Awesomeguy    !!!!

Here's another  method

By Vieta, the product of the roots  =  c

So   c =  

(-5 + sqrt 17) /2   (  -5  -sqrt 17)  / 2   =

(25  - 17)  /  4   =

8 /  4   =  2

The quadratic formula is in format,

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

In this equation, \(x^2 + 5x + c = 0\), a = 1, and b = 5, as we are given that. The only term in the formula that contains c is the discriminant, \(\sqrt{b^2-4ac} = \sqrt{17}\)

Square both sides,

\(b^2-4ac = 17\)

Plug in known values,

\(25-4c = 17\)

Subtract 25 on both sides and divide by -4 to get c = 2

Thanks Heureka,  

I did not read the question properly.   I didn't notice that it was only odd values of r.

f = 9s

in three years

f+3 = 6 (s+3)        sub in the red equation

9s + 3 = 6s + 18

3s = 15

s =   5  y/o

Parallelogram Law      Sum of squares of four sides   = sum of squares of diagonals   ( Thanx, Chris!)

30^2 + 20^2 + 30^2 + 20^2 = 40^2 + x ^2        x = sqrt 1000     you can simplify from here !.....

Their rates per job are   1job/ 3 hours     1/4      and 1/8

   combining their rates for the one job:

1 job / ( 1/3 + 1/4 + 1/8)  = _______    hours     ( multiply this answer by 60 to get 'minutes')

A will do 1/3 of the task in 1 hour. 

B will do 1/4 of the task in 1 hour. 

C will do 1/8 of the task in 1 hour. 

Together, they do 1/3 + 1/4 + 1/8 of the task in 1 hour. 

How many hours does it take?

=^._.^=

What is the tens digit of \(17^{1993}\)?

\(\small{ \begin{array}{|rcll|} \hline \mathbf{17^{1993} \pmod{100}} &\equiv& \left( 17^2 \right)^{996}+1 \pmod{100} \\ &\equiv& \left( 17^2 \right)^{996}*17 \pmod{100} \\ && \boxed{17^2 = 289 \equiv 89 \equiv 89-100 \equiv -11 \pmod{100}} \\ &\equiv& \left( -11 \right)^{996}*17 \pmod{100} \\ &\equiv& \left( (-11)^2 \right)^{498}*17 \pmod{100} \\ &\equiv& \left( 121 \right)^{498}*17 \pmod{100} \quad | \quad 121 \equiv 21 \pmod{100} \\ &\equiv& \left( 21 \right)^{498}*17 \pmod{100} \\ &\equiv& \left( 21 \right)^{5*99+3}*17 \pmod{100} \\ &\equiv& \left( 21^5 \right)^{99}*21^3*17 \pmod{100} \quad | \quad 21^5 \equiv 01 \pmod {100} \\ &\equiv& \left( 1\right)^{99}*21^3*17 \pmod{100} \\ &\equiv& 21^3*17 \pmod{100} \\ &\equiv& 157437 \pmod{100} \\ &\equiv& \mathbf{{\color{red}3}7 \pmod{100}} \\ \hline \end{array} }\)

The tens digit is \({\color{red}3}\)

If we look at the units digit of the first couple multiples of 6, we see

6, 2, 8, 4, 0, 6, ...

It repeats after the fifth term, and every set of those 5 numbers sums up to 20, which has a units digit of 0. That means that the multiples up to 600 also has a units digit of 0, as every set of five is equal to 0. We are looking for the sum of multiples smaller than 600, meaning it would go up to 594, but that does not matter as 600 has a units digit of 0. Therefore, the units digit is 0.

If the chance is 7/15 to get an even page number, then there would be 15 pages, as there are 7 multiples of 2.

Thanks for the info i will try to figure it out for more