# Stuff... again... :)

• Nov 9th 2007, 09:07 PM
Rocher
Stuff... again... :)
1. Frank and Ernest start jogging on a 110m circular track. They begin at the same time and from the same point but jog in opposite directions, one at $\frac{8}{3}m$ per second and the other at $\frac{7}{3}m$ per second. How many times will they meet during the first 15 minutes of jogging?

2. Two candles of the same height are lit at the same time.. The first is consumed in four hours, the second in three hours. Assuming that each candle burns at a constant rate, how many hours after being lit was the first candle twice the height of the second?

3. If a:b = 3:4 and a: (b+c)= 2:5, find the value of a:c.

4. $\frac{1}{2}a = \frac{4}{3}b = \frac{5}{6}c$

5. The volume of two boxes are in the ratio 3:8. If the volume of the larger box is $375cm^3$ more than the volume of the smaller box, find the volume of each box.

Express each of the following with denominator 1:

6. $(x^3)^\frac{1}{3}$

7. $((a-b)^4)^\frac{1}{4}$

Simplify / Evaluate

8. $(\frac{x^4y^6}{a^8b^10})^\frac{3}{2}$

Simplify

9. $\frac{(xy^2)^4}{2x^3y^3}$

Solve

10. $9x=27$

Thank you guys.
• Nov 9th 2007, 10:02 PM
earboth
Quote:

Originally Posted by Rocher
...
Express each of the following with denominator 1:

6. $(x^3)^\frac{1}{3}$

7. $((a-b)^4)^\frac{1}{4}$

Simplify / Evaluate

8. $(\frac{x^4y^6}{a^8b^{10}})^\frac{3}{2}$

Simplify

9. $\frac{(xy^2)^4}{2x^3y^3}$

Solve

10. $9x=27$

Thank you guys.

Hello,

to #6, #7:

$(x^3)^\frac{1}{3}=x=\frac{x}{1}$

$((a-b)^4)^\frac{1}{4}=|a-b|=\frac{|a-b|}{1}$
to #8:

$(\frac{x^4y^6}{a^8b^{10}})^\frac{3}{2}=x^{4\cdot \frac32} \cdot y^{6\cdot \frac32} \cdot a^{-8 \cdot \frac32} \cdot b^{-10 \cdot \frac32}$ = $x^6 \cdot y^9 \cdot a^{-12} \cdot b^{-15}$

to#9:

$\frac{(xy^2)^4}{2x^3y^3}=\frac12 \cdot x^{4-3} \cdot y^{2 \cdot 4 - 3}=\frac12 \cdot x \cdot y^5$

to #10:

Are you sure that you can't solve this equation? (Divide both sides by 9 [you do this to get 1 as the coefficient of x because you want to know the value of one x]. For confirmation only: I've got x = 3)
• Nov 9th 2007, 10:32 PM
earboth
Quote:

Originally Posted by Rocher
1. Frank and Ernest start jogging on a 110m circular track. They begin at the same time and from the same point but jog in opposite directions, one at $\frac{8}{3}m$ per second and the other at $\frac{7}{3}m$ per second. How many times will they meet during the first 15 minutes of jogging?
...

3. If a:b = 3:4 and a: (b+c)= 2:5, find the value of a:c.

4. $\frac{1}{2}a = \frac{4}{3}b = \frac{5}{6}c$

5. The volume of two boxes are in the ratio 3:8. If the volume of the larger box is $375cm^3$ more than the volume of the smaller box, find the volume of each box.

...

Hi,

to #1.: The distance between the two joggers increases by 5 m/s. Together they need 22 s to run the full distance of 110 m.
15 min = 900 s. Therefore they meet $\frac{900}{22} \text{ times} \approx 40\text{ times}$

#3.: From the first equation you get: $b=\frac43 a$ . Plug in this term into the 2nd equation:

$\frac{a}{\frac43 a + c}=\frac25$ . After a few steps of simplification you get:

$\frac{7}{15}a = \frac25 c~\iff~\frac ac=\frac25 \cdot \frac{15}{7} = \frac67$

to #4.: Split this "chain of equalities" into 3 equations and solv for a, b and c:

$\left \{\begin{array}{l}\frac{1}{2}a = \frac{4}{3}b \\ \frac{1}{2}a = \frac{5}{6}c \\ \frac{4}{3}b = \frac{5}{6}c\end{array} \right.$ (Remark: There doesn't exist an unique solution! I've got $[a,b,c]=[40k,15k,24k],~k\in\mathbb{Z}$

to #5.:
Let V be the volume of the smaller box. Then you have:

$\frac{V}{V+375}=\frac38$ Solve for V. You should get V = 225 cm³
• Nov 10th 2007, 07:57 AM
Rocher
to #10:

Are you sure that you can't solve this equation? (Divide both sides by 9 [you do this to get 1 as the coefficient of x because you want to know the value of one x]. For confirmation only: I've got x = 3)[/QUOTE]

Ah dammit, stupid Latex. It was 9^x. So x is an exponent.
• Nov 10th 2007, 08:25 AM
Soroban
Hello, Rocher!

Quote:

2. Two candles of the same height are lit at the same time.
The first is consumed in four hours, the second in three hours.
Assuming that each candle burns at a constant rate, how many hours
after being lit was the first candle twice the height of the second?

The first candle is consumed in 4 hours.
. . In one hour, $\frac{1}{4}$ iof the candle is gone.
. . In $x$ hours, $\frac{x}{4}$ of the candle is gone.
In $x$ hours, there will be: $\left(1 - \frac{x}{4}\right)$ of the candle left.

The second candle is consumed in 3 hours.
. . In one hour, $\frac{1}{3}$ of the candle is gone.
. . In $x$ hours, $\frac{x}{3}$ of the candle is gone.
In $x$ hours, there will be $\left(1 - \frac{x}{3}\right)$ of the candle left.

If the first candle is twice the height of the second candle,

. . we have: . $1 - \frac{x}{4} \;=\;2\left(1 - \frac{x}{3}\right)$

Now solve for $x.$

Quote:

3. If $a:b \:= \:3:4$ and $a: (b+c)\:= \:2:5$,
find the value of $a:c$

We have: . $\frac{a}{b} \:=\:\frac{3}{4}\quad\Rightarrow\quad b \:=\:\frac{4}{3}a$ .[1]

And: . $\frac{a}{b+c} \:=\:\frac{2}{5}\quad\Rightarrow\quad 5a \:=\:2b + 2c$ .[2]

Substitute [1] into [2]: . $5a \:=\:2\left(\frac{4}{3}a\right) + 2c\quad\Rightarrow\quad \frac{7}{3}a \:=\:2c\quad\Rightarrow\quad \frac{a}{c} \:=\:\frac{6}{7}$

Therefore: . $\boxed{a:c\:=\:6:7}$

Quote:

8. Simplify: . $\left(\frac{x^4y^6}{a^8b^{10}}\right)^\frac{3}{2}$
We have: . $\frac{(x^4)^{\frac{3}{2}}(y^6)^{\frac{3}{2}}} {(a^8)^{\frac{3}{2}} (b^{10})^{\frac{3}{2}} } \;=\;\frac{x^6y^9}{a^{12}b^{15}}$

Quote:

10. Solve: . $9x=27$ . ??
Since the problem is way too simple, I'll assume that it says: . $9^x \:=\:27$

We have: . $(3^2)^x \:=\:3^3\quad\Rightarrow\quad 3^{2x} \:=\:3^3$

Therefore: . $2x \:=\:3\quad\Rightarrow\quad\boxed{ x \:=\:\frac{3}{2}}$

• Nov 10th 2007, 08:33 AM
Rocher
Quote:

Originally Posted by Soroban
Hello, Rocher!

The first candle is consumed in 4 hours.
. . In one hour, $\frac{1}{4}$ iof the candle is gone.
. . In $x$ hours, $\frac{x}{4}$ of the candle is gone.
In $x$ hours, there will be: $\left(1 - \frac{x}{4}\right)$ of the candle left.

The second candle is consumed in 3 hours.
. . In one hour, $\frac{1}{3}$ of the candle is gone.
. . In $x$ hours, $\frac{x}{3}$ of the candle is gone.
In $x$ hours, there will be $\left(1 - \frac{x}{3}\right)$ of the candle left.

If the first candle is twice the height of the second candle,

. . we have: . $1 - \frac{x}{4} \;=\;2\left(1 - \frac{x}{3}\right)$

Now solve for $x.$

Is x -4??

Also, I have another question xD

$2^x-1=8$ >_< Latex doesn't work... It's 2 to the exponent x-1 = 8. x-1 is the exponent, the base is 2.
• Nov 10th 2007, 10:54 AM
earboth
Quote:

Originally Posted by Rocher
Is x -4??

Also, I have another question xD

$2^{x-1}=8$ ...

Hello,

Since $8 = 2^3$ your equation becomes:

$2^{x-1}=2^3$

Two powers with equal bases are equal if the exponents are equal too:

Therefore: $x-1 = 3~\iff~x = 4$
• Nov 10th 2007, 01:13 PM
topsquark
Quote:

Originally Posted by Rocher
$2^x-1=8$ >_< Latex doesn't work... It's 2 to the exponent x-1 = 8. x-1 is the exponent, the base is 2.

When you are coding the LaTeX, put what you want in the exponent inside a pair of { }:
2^{x - 1} generates $2^{x - 1}$

-Dan