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The letters A, B, C, D, E, and F represent digits and

\(ABC,DEF\) represents a positive six-digit integer.

What is the number ABC,DEF if \(4(ABC,DEF) = 9(DEF,ABC)\)?

.6 a + .8 p = 88
a = 130 - p
so
.6 (130-p) + .8 p = 88
78 - .6 p + .8 p = 88
.2 p = 10
p = 100/2 = 50

smallest value would be ~ 0    when x is very very large

largest would be 1/5   when x = 0

(0, 1/5]     Yo can finish.....

Answer: \(9\)

Solution:

If H=10, then the following is true:

M+A+T=25

T+E+A+M=42

M+2E+T=50

Notice that TEAM contains MAT in it. So you can take out the MAT and replace it with 25.

E+25=42

E=17

Putting this value of E into MEET gives the following:

M+2(17)+T=50

M+34+T=50

M+T=16

Going back to the first one, you can see that MAT contains MT. Substituting the value found for M+T into MAT gives this:

A+16=25

A=\(9\)

And 1 what?? (or is my chromebook being annoying)?

The second one should be easy.  Give it a go and discuss your thoughts here.

Add the two equations together to get

1/x + 1/x = 10

2/x =10

x = 1/5      

then find y     1/ 1/5   + 1/y = 3

                           5 + 1/y = 3

                            1/y = -2

                              y = -1 /2             you can finish ...yah?

Given two real numbers p and q such that 1/p + 1/q = 1 and pq = 9, what is p + q?   

Given                                                 1/p + 1/q  =  1  

Multiply both sides by pq          pq • (1/p + 1/q)  =  pq  

                                                       pq/p + pq/q  =  pq  

                                                          q   +   p     =  pq   

Given                                                            pq  =  9  

Therefore                                                p  +  q  =  9  

see original post.....

Here is how that works

    suppose polynomial is   3x ^2 + 2x -4      when x = 0: 

                                            0   +   0  - 4                     then when x = 1 :

                                            3   +   2   -4                        now subtract    when x = 0 from when x =1

                                              3 + 2 -4    -   -4   = 3+2 = 5       which is the sum of the coefficients !

Hope this helped :)

-5 < 2/3 t -2

-3 < 2/3 t

t > -9/2        you can finish?

When a plane intersects a different plane, we get two more regions, so

f(n) = f(n - 1) + 2(n - 1).

Then

f(1) = 2

f(2) = 4

f(3) = 8

f(4) = 14

f(5) = 22

f(6) = 32

Adding up all the equations and using an AM-GM estimate, we get that the system has no solutions.

Find the value so that the slopes are the same.....that would make them parallel lines which do not intersect....

If x+y=9 and xy=18, what is the value of x^3+y^3?

Hello Guest!

\(x+y=9\\ x=9-y\ |\ x\ insert \\ xy=18\\ (9-y)y=18\\ 9y-y^2=18\)

\(y^2+9y+18=0\\ y=-4.5\pm \sqrt{20.25-18}=-4.5\pm 1.5\\ y_1=-6\ |\ x_1=15\ |\ \color{blue}x^3+y^3=3159\\ y_2=-3\ |\ x_2=12\ |\ \color{blue}x^3+y^3=1701\)

  !

I don't quite understand how you got to this? Can you explain?

f(1) = 2, f(2) = 3, and f(3) = -4           I think it is just      2 * 3 * -4 = ______________

If the equations are the same...there is infinite solutions ...they are the same line...

   multiply bottom equation by  1 1/3

 4x  -  1 1/3 y = 1 1/3 b               then if a = - 1 1/3    and   1 1/3 b = -8      b = -6     the equations become the same

we can use the system of equations to figure it out.

x+y=9,

xy=18.

solve for x in the first equation and plug it in as the x-value in the second equation

x=-y+9.

(-y+9)y=18.

evaluate

-y^2+9y=18,

-y^2+9y-18=0,

from here we can see that y=6 OR 3 (because it would equal 0), meaning x equals 3 OR 6 (whichever isn't y but it doesn't really matter)

now we can plug in these integers into your final expression: x^3+y^3

3^3+6^3 (or 6^3+3^3 but the order isn't important according to the commutative property of addition) is equivalent to 243.

the answer is 243.

Let

f(x) = 3x^2 + 2 if x <= 3

f(x) = ax^2 - x + 11 if x > 3

Find a if the graph of y = f(x) is continuous

Hello Guest!

\(f(3) = 3x^2 + 2=29\\ f(3) = a\cdot 3^2 - 3 + 11=29\\ 9a+8=29\ |\ -8\\ 9a=21\ |\ :9\)

\(a=\dfrac{7}{3}\)

The graph is \(f(x)=3x^2+2\ |\ x\leq3\\ f(x)=\dfrac{7}{3}x^2-x+11\ |\ x>3\)

  !

2x^2 + 6x + 11      do a little re-arranging

2 (x^2 + 3x)    + 11     now, complete the square for 'x'

2 (x+1.5)^2    - 4.5   + 11

2 (x+ 1.5)²  + 6.5                  Now you can see  h = - 1.5

f(x) = -x - 3     if x <= 1       see if we can find an answer here

  0 = -x-3

  3 = -x

    x = - 3       this is  <= 1     so this is ONE answer

Try for the other piece of the piecewise function:

  0 = 1-x/2      for x >1

x/2 = 1

   x = 2       this will work because the 'x' needs ot be >1 for this piece    

so the only answer is    x = -3     and  2       summed = -1