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We realize that adding 2 to the x coordinate and -7 to the y coordinate gives us M. So we simply just add the 2 and -7 to m to get b.

And thus our final answer is (8,-6) is B.

There are 47 values of b that work.

Yes, a rational number can be expressed as  a/b   where a  and b are integers (and b does not = 0)

7.9 = 79/10

a. 80/360 degrees= 8/36=2/9. Then the arc length of PQ must be 2/9 the circumference. The circumference is 2(pi)(10)=20pi. Then the arc lenght is 2/9*20pi=40/9pi.

b. The area must be 2/9 the total area. The area of the circle is 10^2pi=100pi. So the area of the sector formed by POQ is 2/9*100pi=200/9pi.

1 /(a+b) = 4                 and        1/ (a+c)  = 5        and      1/(b+c) = 6

4a+4b=1                                         5a+5c=1                        6b + 6c =1

a = 1/4-b                                    a = 1/5-c                  c = 1/6-b

        1/4-b  = 1/5 -c

         1/4-b  = 1/5 - 1/6+b

         1/4 -1/5+1/6 = 2b

             b = 13/120                  then   a =  17/120        and   c = 7/120

1/ ( 17/120 + 13/120 + 7/120) = x

x = 3.24 hours

Is there a EASIER way to do this problem???? 

*Keep in mind that ALL labels are wrong*

1. We know that the box with only green balls can't be Box 3 since Matt pulled a red ball from there.

2. Box 2 also can't be the box with only green balls because it's labeled GREEN, and the label is supposedly wrong.

3. We have one choice left, Box 1!

Answer: Box 1 

Puedes aclarar tu pregunta?

Off topic post/useless post...

Do you need help on fractions?

DE = 32.

We first denote b as the total number of tickets booked on airline b, we construct the following equation:

b+b+8+2b+5=61

4b+13=61

4b=48

b=12

So on airline b he booked 12, on a he booked 20, and on c he booked 29.

61 tix = B     +      B+ 8       +     2B+5     solve for B      then A = B+8       C = 2B+5

The sum of all possible values of x is 49.

x + y + z = 4/5.

one sixth

aops problem

If g(t)=5t^2-4t+3 then g(t-2) would be 5(t-2)^2-4(t-2)+3

5(t-2)^2-4(t-2)+3

=5(t^2-4t+4)-4t+8+3

=5t^2-20t+20-4t+11

=5t^2-24t+31 is our final answer.

1, -1     is NOT on the eqation   x - 2y+4 = 0     so the lines cannot intersect there !

5 (t-2)²  - 4 (t-2)  +3        expand, collect 'like' terms and simplify.....

22.6 cm^2   is the answer I find.

Solve the linear system of equations

x - 2y + 3z = 2

2x - 3y + 8z = 7

3x - 4y + 13z = 8

please show steps!

Hello Guest!

Equation method

I    \(x - 2y + 3z = 2\ |\ Change\ of\ sign\ when\ changing\ page\\ x=2+2y-3z\)     

II    \(2x - 3y + 8z = 7\ |\ Change\ of\ sign\ when\ changing\ page\\ 2x=7+3y-8z\ |\ divided\ on\ both\ sides\ by\ 2\)  

     \(x=3.5+1.5y-4z\)

 III  \(3x - 4y + 13z = 8\ |\ change\ of\ sign\ when\ changing\ page\\ 3x=8+4y-13z\ |\ divided\ on\ both\ sides\ by\ 3 \)

       \(x=\frac{8}{3}+\frac{4}{3}y-\frac{13}{3}z\)

I = II    \(2+2y-3z=3.5+1.5y-4z\ |\ change\ of\ sign\ when\ changing\ page\\ z=1.5-0.5y\)

II = III  \(3.5+1.5y-4z=\frac{8}{3}+\frac{4}{3}y-\frac{13}{3}z\)

to be continued

  !

(x-2-sqrt7)(x-2+sqrt7)         expand via distributive property,   collect like terms and simplify....

You are quite right!

I've deleted my answer.

We see that 9x-10=6x+x+20

9x-10=7x+20

2x=30

x=15

And so MN=x+20=15+20=35 That is our final answer.

The equation system has no solution.

But if it were called that, there would be a solution.