# Thread: Due tommorrow problem solving!!

1. ## Due tommorrow problem solving!!

1. Albert has $2.70 more than Bert does. Bert Has three times as much money as Chris does. Together, the three friends have$26.22. How much money does Albert have?

2. A square based pyrapid has a volume of 12a^5b^4 units^3 and has a height of 9a^3 units. Determine the side length of the base.

****BTW.. when i was trying #2, I had to use the volume of square based pyramid. If you dont wanna look it up, it is:

V=2bs+b^2*****

****Another BTW.. if you dont know already, ^=Exponent**** Thanks a million!

2. Originally Posted by princevisram
1. Albert has $2.70 more than Bert does. Bert Has three times as much money as Chris does. Together, the three friends have$26.22. How much money does Albert have?

2. A square based pyrapid has a volume of 12a^5b^4 units^3 and has a height of 9a^3 units. Determine the side length of the base.

****BTW.. when i was trying #2, I had to use the volume of square based pyramid. If you dont wanna look it up, it is:

V=2bs+b^2*****

****Another BTW.. if you dont know already, ^=Exponent**** Thanks a million!

1. See if you can follow where I got these equations from...

$A = B + 2.70$

$B = 3C$ so therefore $A = 3C + 2.70$.

$A + B + C = 26.22$

$3C + 2.70 + 3C + C = 26.22$.

Solve for C, then you can work out B and A.

3. ## Thanks but...?

Okay.. Thank you... but would this be the rest of the question?::

3C+2.70+3C+C=26.22
7C+2.70=26.22
7C=26.22-2.70
7C=23.52
7C/7=23.52/7
C=3.36

4. Originally Posted by princevisram
Okay.. Thank you... but would this be the rest of the question?::

3C+2.70+3C+C=26.22
7C+2.70=26.22
7C=26.22-2.70
7C=23.52
7C/7=23.52/7
C=3.36
What do A and B equal?

5. Originally Posted by Prove It
What do A and B equal?

Okay.. So..

Now, Find B

B=3C
B=3(3.36)
B=10.08

Now, Find A

A=B+2.70
A=10.08+2.70
A=12.78

6. Originally Posted by princevisram
1. Albert has $2.70 more than Bert does. Bert Has three times as much money as Chris does. Together, the three friends have$26.22. How much money does Albert have?

2. A square based pyrapid has a volume of 12a^5b^4 units^3 and has a height of 9a^3 units. Determine the side length of the base.

****BTW.. when i was trying #2, I had to use the volume of square based pyramid. If you dont wanna look it up, it is:

V=2bs+b^2*****

****Another BTW.. if you dont know already, ^=Exponent**** Thanks a million!

Yes it's correct.

For 2. Are you sure you've written down the volume correctly?

7. Originally Posted by princevisram
[snip]
2. A square based pyrapid has a volume of 12a^5b^4 units^3 and has a height of 9a^3 units. Determine the side length of the base.
[snip]
Is that $12a^5b^4$ and $9a^3$ respectively?

It helps if you learn LaTeX so I can read it...

8. Originally Posted by Prove It
Is that $12a^5b^4$ and $9a^3$ respectively?

It helps if you learn LaTeX so I can read it...

yess!! Thankss! Thats my question! Now can you help me solve it please???! Thanks soo much! PS where can i learn LaTeX?

9. Originally Posted by princevisram
yess!! Thankss! Thats my question! Now can you help me solve it please???! Thanks soo much! PS where can i learn LaTeX?
There are lots of online tutorials and help guides in the LaTeX sub forum here.

Ok, first off, the formula for the volume of a square based pyramid is

$V = l^2H$ where l is the side length of the square and H is the height of the pyramid.

You're told $H = 9a^3$ and $V = 12a^5b^4$.

Using these bits of information we get

$12a^5b^4 = l^2(9a^3)$.

Solve for l.

10. Originally Posted by Prove It
There are lots of online tutorials and help guides in the LaTeX sub forum here.

Ok, first off, the formula for the volume of a square based pyramid is

$V = l^2H$ where l is the side length of the square and H is the height of the pyramid.

You're told $H = 9a^3$ and $V = 12a^5b^4$.

Using these bits of information we get

$12a^5b^4 = l^2(9a^3)$.

Solve for l.
I understand to where you are, but i totaly dont understand how to solve for l.

11. Originally Posted by princevisram
I understand to where you are, but i totaly dont understand how to solve for l.
Divide by $9a^3$, simplify, then take the square root.

12. Originally Posted by Prove It
Divide by $9a^3$, simplify, then take the square root.
Oh my goshh!! I totaly dont understand...! Im sorry..

13. Originally Posted by princevisram
Oh my goshh!! I totaly dont understand...! Im sorry..

What's happened to l to turn it into $12a^5b^4$.

It's been squared, then multiplied by $9a^3$.

So to undo all this stuff that's happened, we do the inverse operations in reverse order.

What's the opposite of multiplying by $9a^3$?

What's the opposite of squaring?

14. Originally Posted by Prove It

What's happened to l to turn it into $12a^5b^4$.

It's been squared, then multiplied by $9a^3$.

So to undo all this stuff that's happened, we do the inverse operations in reverse order.

What's the opposite of multiplying by $9a^3$?

What's the opposite of squaring?
would it be:

$12a^5b^4 = l^2\div9a^3$.

???

15. Originally Posted by princevisram
would it be:

$12a^5b^4 = l^2/9a^3$.

??? / being diveded by
No.

We have

$l^2(9a^3) = 12a^5b^4$.

Divide BOTH sides by $9a^3$.

We then have

$l^2 = \frac{12a^5b^4}{9a^3}$

$l^2 = \frac{4a^2b^4}{3}$.

Now take the square root of both sides to get l.

$l = \sqrt{\frac{4a^2b^4}{3}}$

$= \frac{2ab^2}{\sqrt{3}}$

$= \frac{2\sqrt{3}ab^2}{3}$.

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