power27

The best way to do this problem is to find the equation of the line in slope intercept form.

Find the slope first:

\(\frac{7-11}{1-3}=\frac{-4}{-2}=2\)

Now substitute 2 for the slope in y=mx+b and put one of the points in for x and y, I'll use (1,7)

7=2(1)+b

Solve for b, the y intercept

7=2+b subtract 2

5=b so the y-intercept is the point (0,5)

Since ruben was 30 years old last year, this year he is 31 years old. Ruben is 5 years older that Sarah so 31-x=5. Solving this equation gets Sarah is 26 years old right now.

3. Each number in this form is the product of 5 consecutive intigers.

Every number will have at least 2 multiples of 2 multiplied, 1 multiple of 3, 1 multiple of 4 and 1 multiple of 5. Since every number will have numbers multiplied together with these factors, the GCD is 2^3*3*5= 120.

The discriminant for a quadratic is the part of the quadratic formula that is under the square root.

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

\(b^2-4ac\) is the formula for discriminant.

In your quadratic, a=1, b=11 and c=36. Plug these numbers in.

\(11^2-4(1)(36)=-23\)

If the discriminant is negative, the quadratic will have 2 non real solitions. It it is positive, there will be 2 real solitutions. If it is 0 there will be 1 real solitution.

The discriminant is negative so your quadratic will have 2 non real solitutions.

That wouldn't be divisible by 3, but there would always be another combination of 3 numbers in the set that would be divisible by 3.

\(123456789^2-2*123456789*123456787+123456787\)

x=123456789

Rewrite this as:

x^2-2x(x-2)+x-2

(123456787 is 2 less that 123456789 so it can be written as x-2)

Now simplify the equation:

x^2-2x^2+4x+x-2         distribute -2x

-x^2+5x-2                    combine like terms

That can't be factored so I guess that would be as simple as you can get it.

Doing the problem this way would be marginally more efficient, but I still wouldn't want to do that by hand. The problem looks like it is set up to be solved how I solved it, but the numbers didn't work out very well.

Let x be a multiple of 3.

There are 3 possible remainders for a number divided by 3: 0, 1 and 2. Since x is a multiple of 3 the numbers x+0, x+1, and x+2 will leave a remainder of 0, 1 and 2, respectivly, when divided by 3.

Lets look at the sets of numbers that will sum to a multiple of 3:

x, x, x

The sum is 3x which will always be a multiple of 3 because you are multiplyng by 3 and x is a multiple of 3.

x, x+1, x+2

Adding these numbers gets 3x+3, which will always be a multiple of 3.

x+2, x+2, x+2

The sum is 3x+6 which will be a multiple of 3 because both parts of the number are divisible by 3.

x+1, x+1, x+1

Sums to 3x+3, always a muliple of 3 for the above reasons.

It is impossible to have a set of numbers that does not have one of these cases in it.

If the set has one of each remainder, you can sum them to be a multiple of 3.

Lets try to make a set of numbers where none sum to a multiple of 3:

x, x, x+2, x+2

If you have two of 2 different remainders, if you add the third remainder you will have one of each and a multiple of 3. If you add one of the remainders you already have, you will have 3 times that remainder which will always be a multiple of 3.

Oh oops thanks for catching that you can't do that.

\(\frac{1}{\sqrt{2}+\frac{1}{\sqrt{8}+\sqrt{200}+\frac{1}{\sqrt{18}}}}\)

First, simplify the radicles in the expression

\(\sqrt{8}=\sqrt{2 \cdot 2\cdot 2} =2\sqrt{2}\)

\(\sqrt{200}=\sqrt{10\cdot20\cdot2}=10\sqrt{2}\)

\(\sqrt{18}=\sqrt{3\cdot3\cdot2}=3\sqrt{2}\)

\(\frac{1}{\sqrt{2}+\frac{1}{2\sqrt{2}+10\sqrt{2}+\frac{1}{3\sqrt{2}}}}\)

The two radicles can be added together because they have the same base

\(\frac{1}{\sqrt{2}+\frac{1}{12\sqrt{2}+\frac{1}{3\sqrt{2}}}}\)

Now start simplifying from the innermost fraction

Find a common denominator to add the two on the bottom

\(\frac{1}{\sqrt{2}+\frac{1}{\frac{73}{3\sqrt{2}}}}\)

\(\frac{1}{\sqrt{2}+\frac{3\sqrt{2}}{73}}\)

Find a common denominator again

\(\frac{1}{\frac{73\sqrt{2}}{73}+\frac{3\sqrt{2}}{73}}\)

Add the fractions

\(\frac{1}{\frac{76\sqrt{2}}{73}}\)

\(\frac{73}{76\sqrt{2}}\)

Multiply by \(\frac{\sqrt{2}}{\sqrt{2}}\) to get the radicle out of the denominator

\(\boxed{\frac{73\sqrt{2}}{152}}\)

A

mean=\(\frac{A+B+C}{3}\)

For this to be an intiger, A+B+C must be a multiple of 3.

To get the largest possible mean, the numbers would have to be 47, 48 and 49 are A, B and C, respectivly.

47+48+49 = 144 which is a multiple of 3 so I don't need to try anything else (although the sum of any set of 3 consecutive intigers would be a multiple of 3).

144/3 = \(\boxed{48}\) is the greatest possible mean.