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Using the sum of cubes factorization on $a^3+b^3=45$ yields $(a+b)(a^2-ab+b^2)=45$, which gives $7(a^2-ab+b^2)=45$, which gives $a^2-ab+b^2=\frac{45}{7}$

Squaring $a+b=7$ gives $a^2+2ab+b^2=49$. We now have a system of equations! From here, there are two ways to proceed. One is to solve the system of equations, find the values of a and b, and plug them in. The other way, which is MUCH faster, will be shown here.

Subtracting the first equation from the second yields $2ab-(-ab)=\frac{298}{7}$, which gives $ab=\frac{298}{21}$

Now, we make use of the fact that $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$(you can check this yourself by splitting up the fractions), and we plug in! We now have $\frac{a+b}{ab}=\frac{7}{\frac{298}{21}}=\boxed{\frac{147}{298}}$

Sorry, but this was incorrect. Please show ALL YOUR WORK so I can check for errors. And also, yes, the problem asks to find x

Please show your work, how did you even get that answer?

The problem is to find the probability that x < y.  The answer is 1/4.

The x coordiante of the midpoint would be \(\frac{2-(-6)}{2}=\frac{8}{2}=4\) and the y coordinate would be \(\frac{5-3}{2}=\frac{2}{2}=1\) so the midpoint is (4,1) and the product of the coordinates would be 4 * 1 = 4

$\alpha^2 + \beta^2 = \boxed{5}$

The first term is 45/15 = 3.

Hey there, Guest!

So, first, we need to distribute the negative sign:

\(=9x^5+25x^3−4+−1(x^5−8x^3−4)\)

\(=9x^5+25x^3+−4+−1x^5+−1(−8x^3)+(−1)(−4)\)

\(=9x^5+25x^3+−4+−x^5+8x^3+4 \)

Then, we can combine like terms:

\(=9x^5+25x^3+−4+−x^5+8x^3+4\)

\(=(9x^5+−x^5)+(25x^3+8x^3)+(−4+4)\)

\(=8x^5+33x^3\)

In case you missed that, the answer is:

\(=8x^5+33x^3\)

Hope this helped! :)

( ﾟдﾟ)つ Bye

The prime factoriztion of 112 is 2^4*7, so we can factor out 2^3 and get \(\sqrt[3]{112}=2\sqrt[3]{\frac{112}{2^3}}=2\sqrt[3]{14}\). So the answer would be 2 + 14 = 16.

The probability is 28/33.

abc = 48.

(9x^5+25x^3- 4) - (x^5 - 8x^3 - 4) = (9-1)x^5 + (25+8)x^3 + (-4+4) = 8x^5 + 33x^3

13a^3 + 87a^2

(a)( a) ( 13a + 87)

Hey there, Guest!

So, to factor an equation is quite simple, I'll lead you through with these steps:

First, \((6a^3+92a^2-7)-(-7a^3+5a^2-7)\).

Then, 13a^3+87^a2.

(Sorry, LaTeX wouldn't format that properly, but the two is also part of the exponent.)

Lastly, \(a^2(13a+87)\).

And there you have it, your answer.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Shananigans! 

So for this equation, you cannot further simply.

If you're asking us to factor the equation, then I can do that because all of the numbers are divisible by two.

So, the answer would be:

\(2(x−1)(x−2)\)

Hope this helped! :)

( ﾟдﾟ)つ Bye

Use distance formula between these points to find length of one side of the square.....then multiply by 4 sides

sqrt( 5^2 + 12^2 ) =   13 units  side length   .....you can finish....

Expand L side      add to R side to get

9x^2 + 36x + 32    factor to get

(3x+8)(3x+4)

There are 284 ways to choose the three chairs.

Not what the question asked for ....but nice solution!          If you just drop the '6' from your final answer you will get  3:8 as the answer...Lev to Naomi ages.

Maybe you meant 'factor' it ?

(2x-4)(x-1)

Hey there, Chese!

This question on ratios is quite simple; let me demonstrate below.

Lev:Mina is 1:2
Mina:Naomi is 3:4
Lev:Mina is 3:6 and Mina: Naomi is 6:8
Therefore, Lev:Mina:Naomi is 3:6:8

Hope this helped! :)

( ﾟдﾟ)つ Bye

Let my try posting it again...

Points $A$,$B$,$C$,$D$, are located on $\overline{AB}$ such that $AB=3AD=6BC$. If a point is selected at random on $\overline{AB}$, what is the probability that it is between $C$ and $D$? Express your answer as a common fraction.

Better?