# Some questions!

• Sep 5th 2007, 11:18 AM
Xplicit
Some questions!
1) Make p The subject of the formula below

q= 1/5p=2
(Thats 1 over 5)

2) v = fw

Calculate the value of v when
f=4x10 to the par of 4
w=5.7x10 to the par -7

3) Calculate the value of f when

v = 2.8x10 to the par of 8
w = 5.7x10 to the par -7

4) Mrs brown paid £115.20 for parts and £38.40 for labour

What % of the cost was for labour?

5) Solve the following inequality where X can take any value
-12 < (Or equal to) 3X < 9

6a) Round the following numbers to 1 signifiant figure your answer will be an approximation

11.8 x 3.41
-----------
5.52

6b)

0.0029 x 0.041 squared

7) Solve the simultaneous equations

7x - 4y = 37
6x + 3y = 51

8) Not sure if you can do this on a forum but o well worth a try

On the grid below, leave unshaded the region satisfying the inequalities

-2 < ( or equal to) X < (or equal to) 2
0 <(or equal to) Y < (or equal to) 3

graph goes from the x axis -4 to 4
Y axis is -4 to 4

• Sep 5th 2007, 12:25 PM
topsquark
Quote:

Originally Posted by Xplicit
1) Make p The subject of the formula below

q= 1/5p=2
(Thats 1 over 5)

2) v = fw

Calculate the value of v when
f=4x10 to the par of 4
w=5.7x10 to the par -7

3) Calculate the value of f when

v = 2.8x10 to the par of 8
w = 5.7x10 to the par -7

1) Which do you want?
$\displaystyle q = \frac{1}{5}p \implies p = 5q$
or
$\displaystyle \frac{1}{5}p = 2 \implies p = 2 \cdot 5 = 10$

2) (Chuckles) What is "par?" The British accent version of "power?" :D
$\displaystyle v = fw = (4 \times 10^{4})(5.7 \times 10^{-7})$

Rearrange things a bit...
$\displaystyle v = (4 \cdot 5.7) \times (10^4 \cdot 10^{-7})$

$\displaystyle v = 22.8 \times 10^{4 - 7} = 22.8 \times 10^{-3}$

Now shift the decimal:
$\displaystyle v = 2.28 \cdot 10^1 \times 10^{-3}$

$\displaystyle v = 2.28 \times 10^{1 - 3}$

$\displaystyle v = 2.28 \times 10^{-2}$

3. This is the same method as the above. Try it. For comparison I get $\displaystyle 1.596 \times 10^2$.

-Dan
• Sep 5th 2007, 12:35 PM
topsquark
Quote:

Originally Posted by Xplicit
4) Mrs brown paid £115.20 for parts and £38.40 for labour

What % of the cost was for labour?

5) Solve the following inequality where X can take any value
-12 < (Or equal to) 3X < 9

6a) Round the following numbers to 1 signifiant figure your answer will be an approximation

11.8 x 3.41
-----------
5.52

6b)

0.0029 x 0.041 squared

4) Percent cost of labour is going to be:
$\displaystyle \frac{\text{cost of labour}}{\text{total cost}} \times 100 \%$

5)
$\displaystyle -12 \leq 3X < 9$

Split this into two inequalities. (You can also do this stuff all at once, but since you are fairly new to this I'll try to keep it simple.)
The first half:
$\displaystyle -12 \leq 3X$

$\displaystyle \frac{-12}{3} \leq X$

$\displaystyle -4 \leq X$

The second half:
$\displaystyle 3X < 9$

$\displaystyle X < \frac{9}{3}$

$\displaystyle X < 3$

Putting the two pieces back together:
$\displaystyle -4 \leq X < 3$

6a) $\displaystyle \frac{11.8 \times 3.41}{5.5} = 7.316$

To one significant figure we round to the ones place. Thus the answer is 7.

6b) $\displaystyle 0.0029 \times (0.041)^2 = 0.0000048749$

Probably the first thing you will want to do here is convert this to scientific notation, so we need $\displaystyle 4.8749 \times 10^{-6}$. Again, we need to round to the ones place (in the scientific notation, of course!) so we get $\displaystyle 5 \times 10^{-6}$. If you insist, we can write this as 0.000005. (Notice that none of the 0's in front of the 5 are significant. They are merely place-holders.)

-Dan
• Sep 5th 2007, 12:44 PM
topsquark
Quote:

Originally Posted by Xplicit
7) Solve the simultaneous equations

7x - 4y = 37
6x + 3y = 51

There are a whole bunch of ways to solve something like this. I'm going to use the substitution method.

Solve the top equation for, say, y.
$\displaystyle 7x - 4y = 37$

$\displaystyle -4y = -7x + 37$

$\displaystyle y = \frac{-7x}{-4} + \frac{37}{-4}$

$\displaystyle y = \frac{7x}{4} - \frac{37}{4}$

Now insert this value of y into the second equation:
$\displaystyle 6x + 3y = 51$

$\displaystyle 6x + 3 \left ( \frac{7x}{4} - \frac{37}{4} \right ) = 51$

$\displaystyle 6x + \frac{21x}{4} - \frac{111}{4} = 51$

Let's get rid of those fractions and multiply both sides by 4:
$\displaystyle 24x + 21x - 111 = 204$

$\displaystyle 45x = 204 + 111 = 315$

$\displaystyle x = 7$

Now put this value of x into the y equation (or either of the original equations):
$\displaystyle y = \frac{7 \cdot 7}{4} - \frac{37}{4} = 3$

So the solution is $\displaystyle (x, y) = (7, 3)$.

-Dan
• Sep 5th 2007, 01:00 PM
Xplicit
Just 2 more question
What fraction must be written in the box to make the statement true

7/16 + 1/4 = BOX! x 3/4

2)45000= 90x500
Express 4500 as a product of its prime factors

And thanks ALOT!
• Sep 5th 2007, 01:04 PM
topsquark
Quote:

Originally Posted by Xplicit
Just 1 more question
What fraction must be written in the box to make the statement true

7/16 + 1/4 = BOX! x 3/4

And thanks ALOT!

Well, call the unknown value in the box something like y:
$\displaystyle \frac{7}{16} + \frac{1}{4} = y \times \frac{3}{4}$

$\displaystyle \frac{7}{16} + \frac{4}{16} = y \times \frac{3}{4}$

$\displaystyle \frac{11}{16} = y \times \frac{3}{4}$

$\displaystyle y \times \frac{3}{4} = \frac{11}{16}$ (Just turning it around.)

$\displaystyle y \times \frac{3}{4} \times \frac{4}{3} = \frac{11}{16} \times \frac{4}{3}$

$\displaystyle y = \frac{11}{12}$

-Dan