NotThatSmart

Let's let the two sides of the rectange x and y. 

We can use the problem to write the equation

\(2x + 2y = 40 \\ x + y = 20 \)

Isolating y, we have \(y = 20 - x\). 

Subsituting this value for y into the second part of the qeustion and using the pythaogrean theorem, we have

\(x^2 + y^2 = (10 \sqrt{2}) ^2 \\ x^2 + ( 20 - x)^2 = 200 \\ x^2 + x^2 - 40x + 400 = 200 \\ 2x^2 - 40x + 200 = 0 \\ x^2 - 20x + 100 = 0\)

Finally, we find that we have x = 10. From the first equation, we find that y = 10 as well. 

The area is just 10*10=100!

Thanks! :)

This isn't too hard of a question once you realize how to do it!

\(\Delta EFG \cdot \frac{2h}{3}=18\\ \Delta EFG =\frac{3\cdot 18}{2h}\\ [{EFGH}]=\Delta EFG\cdot \frac{1}{3}h\\ [{EFGH}]=\frac{3\cdot 18}{2h}\cdot \frac{1}{3}h\\ [{EFGH}]=9\)

The key is realizing that the pyramid is just 1/2 of EFG times the height!

Thanks! :)

This isn't too bad of a problem!

First off, we have to find how many numbers between 1 and 1000 are divible by 5. 

It might seem challenging, but what it really is is just 1000/5 = 200! 

Now, there are 1000 numbers between 1 and 1000, so we just have 200/1000, which gets us 1/5. 

So there is just a 1/5 chance that there will be an integer divisble by 5!

Thanks! :)

First, we have to write this in standard form by combining all like terms and moving all terms to one side. We get \(x^2+11x-49>0\)

Using the quadratic equation, we get \(x<\frac{-\sqrt{317}-11}{2}\quad \mathrm{or}\quad \:x>\frac{\sqrt{317}-11}{2}\). 

Putting this into interval notation, we have \((\frac{-\sqrt{317}-11}{2}, \frac{\sqrt{317}-11}{2})\). 

This isn't a hard problem to do. It is annoying we have square roots and an unfactorable polynomial, but not too bad in general!

Thanks! :)

First, let's find how many ways we can choose 3 randomly, no restrictions. 

Using the glorious formula for combinations, we have \(9 \choose 3\)\(= 84\). This means there are 84 ways to choose 3 circles randomly. 

Now, let's see how many DON'T work. 

There are 6 rows and columns of 3 circles, and each of these adjacent 3 have 3 ways we can have an inablid combination. 

Thus, we have 6*3 = 18. 

\(84-18=66\). There are 66 ways we can do this!

I'm not really sure if this is correct...

Thanks! :)

I graphed it using a graphing calculator by graphing each side of the equation

The intersection gets us something like\(x \approx 11.18119077\)

I don't think this was the exact answer you were looking for, but my brain isn't smart enough to solve this ;)

Thanks! :) 

Let t be the time it takes to drive to the beach. We have

\(40\ mph\cdot t=55\ mph(t-25\ min)\\ 40t=55t-1375\ min\\ 15t=1375\ min\\ t=91\frac{2}{3}\ min\)

Now, we know that \(s=v\cdot t\) from our speed formulas. We just have to plug the values in!

We have 

\(s=v\cdot t=40\ mph\cdot 91\frac{2}{3}\ min\cdot \frac{h}{60\ min}\\ s=61 \frac{1}{9}\ miles\)

Thanks! :)

This question isn't hard at all, and only involves one key fact. 

First, we HAVE to know the formula 

Dilation Factor = Length of Image / Length of Pre-image

So, now, we just plug in some numbers and we are done!

Length of AB (image) = 30 units

\(5 = 30 / OA \)

\(OA = 30/5\)

\(OA = 6\)

So OA is basically just 6 units. 

Thanks! :)