NotThatSmart

Let adjacent  sides be x, y

Thus, we have the equation

2x + 2y  =  40

x + y  = 20

y = 20 - x

By the Pythagorean Theorem

x^2  + y^2  = (10 sqrt 2) ^2

x^2 + ( 20 - x)^2  =  200

x^2 + x^2  - 40x + 400 =  200

2x^2  - 40x + 200  = 0

x^2 - 20x + 100  = 0

(x -10)^2  = 0           take positive root

x  - 10  =  0

x = 10

y = 20 - 10  =  10

Area = xy =  10 * 10  =   100

I have respect for you Answerscorrectly. 

It's easy just to copy off answers here. I know I sound like a teacher right now, but seeing how I KNOW most users are here just to see answers, I deeply respect you asking for hints rather than going for the easy way. 

Hats off to you! :)

Also, good luck in that C and P course. I took that 2 years ago. 

All I have to say is...ummm....it....umnmm....I struggled, but I learned a lot. 

:)

I know this is stupid, but I'm kinda dumb, so here's the entire polynomial

(This took a while)

\(729y^{12}+2916y^{11}-2430y^{10}-19980y^{9}+135y^{8}+59976y^{7}+6364y^{6}-99960y^{5}+375y^{4}+92500y^{3}-18750y^{2}-37500y+15625\)

The answer is 375...

Thanks! :)

We can use subsitution to our advantage. 

Now, first, let's put a terms of b. From the second equation, we get that

\(a=\frac{4}{b^2}\)

Subsituting this value of x into the first equation, we get

\(\frac{4}{b^2} \cdot b^4 = 48\\\)

Now, the top and the bottom cancel out, so we have

\(4b^2 = 48\\ b^2=12\\ b=2\sqrt{3},\:b=-2\sqrt{3}\)

Thus, we find two values of b. We have

\(b=2\sqrt{3},\:b=-2\sqrt{3}\)

Thanks! :)

We could isolate a from the first equation and get \(a=\frac{250}{1+b}\) before plugging it into the second equation, but let's do something simpler. 

Adding the first and the secone equation, we get

\(2a=10\\ a=5\)

Plugging this in for b, we get

\(a=5,\:b=49\)

So, our final answer is 5. 

Thanks! :)

*1200 points

First, let's isolate a^3 as much as possible. 

We have

\(\frac{1}{64a^3+7}=7\\448a^{3}+49=1\\ 448a^3=-48\\ a^3=-48/448 \)

From this, we get that

\(a=\sqrt[3]{-28/3}\)

This is actually the only real factor for a, so it is our final answer. 

Thanks! :)

First, moving all terms to one side other than the constants, we get

\(x^2+y^2-6x-6y=56\)

Completing the square for both x and y, we get

\(\left(x-3\right)^2+\left(y-3\right)^2=\left(\sqrt{74}\right)^2\)

Thus, the radius is

\(r=\sqrt{74}\)

Using the circle equation, we have

AREA OF CIRCLE = \(\sqrt{75}^2\pi = 74\pi\)

Thus, 74pi is the answer. 

Thanks! :)

First, let's simplify the equation we are given. We get

\(10x^2 + 10x - 1\\ ab = -1/10\\ 2ab = -2/10 = -1/5\)

Now, take a + b  =  -1

Squaring both sides, we get

\(a^2 + 2ab + b^2  = 1 \\ a^2 - 1/5 + b^2  = 1 \\ a^2 + b^2  = 1 + 1/5  =  6/5\)

\((a -4) / ( b -4) + ( b -4) / (a -4) = [ (a - 4)^2 + ( b -4)^2] ] ' [ (a -4) ( b-4) ] = \\ [ a^2 - 8a +16 + b^2 - 8b + 16 ] / [ ab - 4(a + b) +16 ] = \\ [ (a^2 + b^2) - 8(a + b) + 32 ] / [ ab - 4 (a + b) + 16 ] =\\ [ 6/5 - 8 (-1) + 32 ] [ -1/10 - 4 (-1) + 16 ] = \\ [ 6/5 + 40 ] / [ 20 - 1/10] = 20497 / 25\)

The answer is \(20497 / 25\)

Thanks! :)

Simplifying the equation, we have \(x^2 + 6x - 2  = 0 \)

In the form mx^2 + nx + p  = 0, the Product of the roots ab = -n/m =   -2  

So 2ab = -4

Sum of the roots a + b =  p/m =  -6     

Squaring both sides, we have 

\(a^2 + b^2 + 2ab = 36 \\ a^2 + b^2 - 4 =36 \\ a^2 + b^2  = 40 \)

\(1/a^3  + 1/b^3  = \\(a^3 + b^3) / (ab)^3   \)        Note  :  a^3 + b^3 = (a + b) (a^2 + b^2 - ab)

So we have

\((a + b) ( a^2 + b^2 -ab)  / (ab)^3  = \\ (-6) (40 - -2) / (-8)   =  \\ (6) (42 / 8)  =   31.5 \)

Thus, the answer is 31.5

Thanks! :)

First, let's calculate the distance Will travels before grace even leaves. 

Since he traveled for 45 minutes, we have

\((45 min)(50 m/min) = 2250 m \)

So when Grace starts, she and Will have

\((2800 m) – (2250 m) = 550 m \) to cover. 

Let's let t be the amount of time it takes for the two to catch up to each other after Grace starts rowing. 

We have the equation

\( (50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875 \)

This is approximately 6 minutes and 52 seconds. 

So the clock time they meet is 6 min 52 sec after Grace starts.

Grace started at 2:45, so add 6 min 52 sec and the clock will read 2:51:52 pm when they meet  

Thus, our final answer is \(2:52:52\)

Thanks! :)