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To make things easy, let  the box  length = 6  and the   box width =10  and the box  height  = 5

We can put 3 cubes along the length  and 5 cubes along the width.....thus....each layer of cubes contains 3 x 5 =  15 cubes

And we  can  stack  2  layers of cubes  =  2x15 = 30 cubes in total

The volume  of  the box is  6 * 5 * 10 =  300 in^3

The volume of the  cubes =  2^3 * 30  = 240 in^2

% of box volume taken up  by the  cubes  =  240 / 300  =  8/10  = 80%

Thank you! 

Thanks so much Spring!

Thanks so much, CPhill 

So long as I have helped you to learn, I am pleased.    

5(x + 10)  =250     divide  both sides  by 5

x + 10  = 50          subtract 10 from both sides

x =  40

The expression or equation is: 5x-10=20

First, add 10 to both sides: 

5x-10+10=20+10

5x=30

Then, divide:

5x/5=30/5

x=6

So, 5 times 6 =30.

Subtract 10 from 30 =20.

Thanks so much!

First, divide both sides by 2:

2(x+8)/2=24/2

x+8=12

Subtract 8 from both sides:

x+8-8=12-8

x=4

Wow! Thank you so much Melody. I greatly appreciate it! 😊

yes so it will be

e^(i(5pi/6)  = cis 5pi/6

Next, you subtract 7π/6 & π/3. When you subtract you get 5π/6.

right!

x^7 divided by x^2 = x^(7-2)= x^5

When you do not understand what to do always convert to a simply but similar example and see what happens.

\(cis(\theta) = e^{i\theta}\\~\\ cis(7\pi/6) \div cis(\pi/3)\\= e^{i(7\pi/6)}\div e^{i(\pi/3)}\\ =e^{[(i(7\pi/6)-i(\pi/3)]}\\ =e^{i[(7\pi/6)-(\pi/3)]}\)

What is next?

Okay, thank you.

You subtract 7-2 to get 5. The answer is x^5.

Yes the number out the front is correct.

How do you do       x^7 divided by   x^2   ?

Please answer quickly.

Well, subtract 25 from both sides, then simplify by dividing.

25-x=5

25(-25)-x=5-25

-x=-20

-x/-1=-20/-1

x=20

Hope it helps! :)

Well, to be quite honest, I am not 100% sure. I didn't learn it in class with the equation you provided me. 

So solving it to my ability with your equation, correct me if I am incorrect, I got 5π/6.

Also, is the 2√15/5 correct?

Thank you Melody for your time and patience in helping me figure this out. I am not the best at Pre-Calc/Trig! :)

\(cis(\theta) = e^{i\theta}\\ cis(7\pi/6) \div cis(\pi/3)= e^{i(7\pi/6)}\div e^{i(\pi/3)}=etc\)

No it is not.

What is theta?

What is alpha?

What is theta-alpha ?

Find the distance between \(P_1\)(3,4) and the line 4x+8y+7=0.
 

Hello Guest!

\(f_1(x)=y=-\frac{1}{2}x-\frac{7}{8}\)

\(m_1=-\frac{1}{2}\\ {\color{blue}m_2=-\frac{1}{m_1}}=-\frac{1}{-\frac{1}{2}}\\ \color{blue}m_2=2\)

\(\color{blue}f_2=y=m_2(x-x_p)+y_p\\ y=2(x-3)+4\\ y=2x-2\)

\(-\frac{1}{2}x-\frac{7}{8}=2x-2\\ \frac{5}{2}x=\frac{9}{8}\\ \color{blue}x_2=0.45\\ \color{blue}P_2\\ \color{blue}y_2=-1.1 \)

\(\color{blue}D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\ D=\sqrt{(-1.1-4)^2+(0.45-3)^2}\\\)

\(D=5.702\)

The distance between \(P_1\)(3,4) and the line 4x+8y+7=0 is D = 5.702.

  !

I got 7π/6. Is that correct?

You are welcome.  

Hint:

\(cis(\theta) = e^{i\theta}\)

Using this it is relatively easy to determine the answer.   

As  \(e^{i\theta}\div e^{i\alpha}=e^{i(\theta -\alpha)}\)

I think it is 5

Hi Ginger, it's so good to hear from you again.  Thanks for the link to the helllllppppp_2 thread, I noted with delight your comment that "dumbness is contagious."  It reminded me of two things.  I'll start with the shorter:  Insanity is hereditary .... you get it from your children.  Parents will know what this means. 

The other thing is a concept that was discussed on the radio show Car Talk, the question whether two people together can know less than either of them individually.  The two hosts on the show were known for self-deprecating humor, and the question was submitted by a fan of the show.  The discussion ran for weeks, but if it was ever resolved, I missed that episode. 

I almost adopted the username MisterBB for this site, in your honor, but I checked it out first on the internet and discovered to my disappointment that it had already been claimed.  Sigh... gang aft agley. 

Goodnight, Gracey.  

_.