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Sum of cubes => a^3 + b^3 = (a + b)(a^2 - ab + b^2)

a^2 - ab + b^2 = (a + b)^2 - 2ab - ab

a^3 + b^3 = 14((14)^2 - 3ab)

812/14 = 196 - 3ab

58 - 196 = -3ab

ab = 46

a + b = 14.

By vieta's formula, we can write a quadratic: x^2 - 14x + 46 = 0, where a and b are the solutions to the quadratic.

[14 +- sqrt(196 - 184)]/2

(14 +- sqrt(12))/2

(14 +- 2sqrt(3))/2

7 +- sqrt(3).

Thus (a, b) = \((7 + \sqrt{3}, 7 - \sqrt{3}) \) and you can reverse a and b in the ordered pair.

Thanks Ginger, it makes more sense now.

4) You did not fully fill out the question --- please re-post/edit the bottom question since it got cut off after the "Find".

3) In the diagram below, we have.....

Since AB || CD, by alternate interior angles, angle AEG = angle EGF = x*.

Also by alternate interior angles, angle BEF = angle EFG = 102 + x*.

Also EF = FG so the triangle within the two parallel lines is isosceles, by Isosceles Triangle Theorem, angle GEF = angle EGF = x.

The interior angle sum of a triangle is 180 degrees, so 102 + x + x + x = 180. 3x = 78, x = 26.

2)

Area of triangle AXY now has area "x".

Since all the triangles in the image share the same height (the perpendicular from apex A to WZ), then the ratio of the triangles' areas are the ratio of their bases. WY/XZ = 12/10, so the areas of triangles AWY/AXZ = WY/XZ = 12/10. 

Substituting in, we have 15 + x / 8 + x = 12/10 

150 + 10x = 96 + 12x

54 = 2x

x = 27. Triangle AXY has area of 27 units squared.

1)

ADE and CDE share the same height, but different bases, given the ratio of areas being 4 : 6 (ADE : CDE), we can conclude that the ratio of AD:DC is 2:3

Also, the area of triangle ABD is 14 + 4 = 18 units squared. 

The triangles ABD and BCD share the same height, so the ratio of their areas is the ratio of AD:DC, which is 2:3, then we can conclude that triangle BCD has area of 27 units squared. Subtracting 6, the area of triangle BCE is 21 units squared.

ac - 3 and bc - 3 are perfect squares. (also i am assuming that c is an integer cuz duh)

4c - 3 and 7c - 3 are perfect squares

4c - 3 = p^2, 7c - 3 = q^2

Let's do the 4c - 3 first: p^2 must be odd because when you add 3, it gets even and only evens can be divided by 4.

p^2 + 3 is divisible by 4.

p^2 must be odd, so the squared can be 1, 3, 5, 7 (note how all these +3 are divisible by 4), 9, etc.

So 4c - 3 = p^2 doesn't really help. All we can conclude is that c >= 1 and is an integer.

7c - 3 = q^2: q^2 + 3 is divisible by 7. Yowch, divisiblity of 7 is annoying... q = 2 works, 5 works, 9 works, 12 works, ...

To find more solutions, just make sure that q^2 + 3 is divisible by 7, find the possible values of q, and then use algebra to deduce c... There are a bunch of different solutions (so long as c >= 1 and is an integer).

40% of the goals are scored by the star player, that means 40% of the 70 goals are scored by the star player during the season.

40% = 40/100 (remember this) = 0.4. 

0.4 x 70 = 28

70 - 28 = 42.

The rest of the team scored 42 points in total during the season.

Let S=1+3+5+...+(2n-3)+(2n-1)

Then S=(2n-1)+(2n-3)+...+5+3+1

Since S has n terms, and if you add the corresponding terms, the sum is 2n, we have 2S=2n*n, so S=n^2.

I hope this helped!

product_(k=2)^15 (k - 1)/(k + 1) = 1/120

53 is ther answer

and it's cut off in the title, it's supposed to be: when the same constant is added to the numbers $60,$ $100,$ and $160,$ a three-term geometric sequence arises. What is the common ratio of the resulting sequence?

The anwser is 1, 1*4-3=1^2 and 1*7-3=2^2

Also, please help with this:

I choose three different numbers out of this list and add them together: 1, 3, 5, 7, 9,..., 105​. How many different totals can I get? 

After we put 1 of each action figure in each box, there are 3 Scuba Steves and 2 Diving Dans left.

Using stars and bars for the number of Scuba Steves gives us 3 stars and 3 bars, for \({6 \choose 3} = 20\) combinations

Doing the same thing for the Diving Dans gives us 2 stars and 3 bars for \({5 \choose 3} = 10\) combinations.

So, there are \(20 \times 10 = \color{brown}\boxed{200}\) ways to do this.  

no typo, it came with the problem

the anwser should not repeat, for example 105+1+3 and 103+1+5 have the same anwser.

ps: I think that anwser is 150, can anyone check?

[105  -  1] / 2  +  1 ==53 terms

53 C 3==23,426 totals.

You have not responded to your last answers yet.  

Some good manners can go a long way. 

The first two are correct, but you need to look at the complex roots again.

the number of ordered pairs (m,n)

Hello Guest!

\(\color{blue}3/2n = 4-2m\\ 2n=\dfrac{3}{4-2m}\\ n=\dfrac{3}{8-4m}\\ \color{blue}4(m-n) = 7(1-n)\\ 4m-\dfrac{12}{8-4m}= 7-\dfrac{21}{8-4m}\\ \dfrac{32m-16m^2-12}{8-4m}=\dfrac{56-28m-21}{8-4m}\)

\(-16m^2+32m+28m-12-56+21=0\\ 16m^2-60m+47=0\\ \text{a_______b_____c}\\ m = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ m\in\{\frac{15-\sqrt{37}}{8},\frac{15+\sqrt{37}}{8}\}\\ \color{blue}m\in \{1.11465,2.63534\}\)

\(n=\frac{3}{8-4m}\\ n\in \{\frac{3}{8-4\cdot \frac{15-\sqrt{37}}{8}},\frac{3}{8-4\cdot \frac{15+\sqrt{37}}{8}}\}\\ \color{blue}n\in \{0.847127,-1.18046\}\)

  !

twistedwhyami,

Guest spent time looking at your question and asked you to verify the information.

Instead of thanking guest for their time and verifying the info, you just gave a thumbs down.

Do you really think this was an appropriate response. 

Why would anyone bother trying to help you if you are abusvie in return.

fyi   I also think that you should verify the info in your question.

How many books were there?

How much money did each one cost?

multiply.