Power fractions

Problem:



$ \frac{12^9}{12^2} $

Solution:

When you divide, power subtracts, so think of

$ \frac{12 \times 12 \times 12 \times 12 \times 12 \times 12 \times 12 \times 12 \times 12}{12  \times 12} = 12 \times 12 \times 12 \times 12 \times 12 \times 12 \times 12 = 12^7 $.

Or more succinctly, $ \frac{12^9}{12^2} = 12^{9-2} = 12^7 $.

Concentration

Problem:



Solution:

Let the volume of the pure acid be $ x $ in liter.
Let the volume of the 20% acid be $ y $ in liter.

We have two requirements:

$ x + y = 44 $, we ultimately need 44 liters.
$ x + 0.2y = 44 \times 0.4 $, The acid content should be 40%

We simply subtract the second equation from the first and get

$ (x + y) - (x + 0.2y) = 44 - 44 \times 0.4 $, simplifying, get $ y = 33 $.
Now we have $ x + y = 44 $, so we get $ x = 11 $.

As a sanity check, we compute $ 11 + 33 \times 0.2 = 17.6 = 44 \times 0.4 $, so the answer is correct.

Simplify

Problem:
Simplify: $ \frac{x^2 + 2xy + y^2}{x+y} \cdot \frac{x - y}{4} $
Solution:

$ \begin{eqnarray*} & & \frac{x^2 + 2xy + y^2}{x+y} \cdot \frac{x - y}{4} \\ &=& \frac{(x+y)^2}{x+y} \cdot \frac{x - y}{4} \\ &=& (x+y) \cdot \frac{x - y}{4} \\ &=& \frac{(x+y)(x-y)}{4} \end{eqnarray*} $

Plot?

Problem:

$ f(x)=x+\frac{2}{x}-1 $

Solution:

Without further instruction, I will provide a plot for the function here. For small $ |x| $, the graph is dominated by the $ \frac{1}{x} $ component. Otherwise it is dominated by the linear component. Of course, $ x $ cannot be exactly 0, as it is undefined there.

Function notations

Problem:



Solution:

Part b)

We have $ 1945 - 1950 = -5 $, therefore, the answer is $ w(-5) $.

Part c)

We have $ 1982 - 1950 = 32 $, therefore the average number of acres per farm in 1982 is w(32)
We have $ 1930 - 1950 = -20 $, therefore the average number of acres per farm in 1930 is w(-20)

The change (i.e. answer) is $ w(32) - w(-20) $.

Complex Quadratic

Problem:



Solution:

We will use the "completing the square" method to solve this problem.

First. note that $ (x = 1)^2 = x^2  - 2x + 1 $ (by simply multiplying).
We can get

$ x^2 - 2x + 2 =  x^2 - 2x + 1 + 1 = (x - 1)^2 + 1 = 0 $

Rearranging, we get

$ (x - 1)^2 = -1 $.

This implies the solution has no real solution because no real number's square is a negative number. But if you knew complex number, you will recognize $ (x - 1) = \pm i $, where $ i $ is the imaginary unit, so the two complex solutions are $ 1 \pm i $.

Area table

Problem:



Solution:

The radius is half the square side length $ r = \frac{x}{2} $
The square plot area is square of the side length $ A_{square} = x \times x $
The flower area is $ A_{flower} \pi \times r \times r $
The grass area is $ A_{grass} = A_{square} - A_{flower} $.

The numbers are computed as follow:

Square side	Radius	Square Area	Flower Area	Grass Area
1.00	0.50	1.00	0.79	0.21
2.00	1.00	4.00	3.14	0.86
3.50	1.75	12.25	9.62	2.63
6.00	3.00	36.00	28.27	7.73

Leaky bucket

Problem:

Solution:

Bucket A is leaking at a rate $ \frac{\frac{7}{4}}{\frac{3}{2}} = \frac{7}{6} = \frac{21}{18} $
Bucket B is leaking at a rate $ \frac{\frac{5}{6}}{\frac{3}{4}} = \frac{10}{9} = \frac{20}{18} $

Therefore bucket B is leaking slower.

Video rental

Problem:



Solution:

4 video cost (98 - 84) = $14
1 video cost $3.5

The card can be used for 84/3.5 = 24 more videos.

Parallel lines

Problem:




Solution:

130^o

Quadratic function

Problem:



Solution:

Power

Problem:

$ (x - 10)^5 $

Solution:

$ x^5 - 50x^4 + 1000x^3 - 10000x^2 + 50000x - 100000 $

Sequence

Problem:



The value of $ \sqrt{ 6 + \sqrt { 6 + \sqrt { 6 + .... \infty }}} $ is

Solution:

3

Quadratic Graph

Problem:

Solution:

Average change

Problem:

The value of a car changed by -$4,510, -$3,866, -$2,830, -$2,125, -$2, 030, and -$1,964 over 6 years. What is the average change per year?

Solution:

Just compute the average of these numbers, we have

$ \frac{(-4510) + (-3866) + (-2830) + (-2125) + (-2030) + (-1964)
}{6} = \frac{-17325}{6} = -2887.5 $

Table lookup

Problem:

 

Solution:

Part a) When v change from -1.08 to 1.52, we see b goes from 9 to 5.17, so the change is 5.17 - 9 = -3.83

Part b) When b change from 9 to -1.63, we see v goes from -1.08 to 7.27, so the change is 7.27-(-1.08) = 8.35

Part c) When v change from 7.14 to 7.27, we see b goes from -1.08 to -1.63, so the change is -1.63 - (-1.08) = -0.55

Jameson

Problem:

Jameson must divide 167 gallons of used oil up between three barrels of varying size. He puts in the first barrel one and a half times as much oil as he does the second barrel. He puts in the first barrel 15 gallons less oil in the third barrel. How much oil does he put in each barrel?

Solution;

Let $ x $ , $ y $ and $ z $ be the number of gallons of oil in the first, second and third barrel respectively.

The first condition is this:
He puts in the first barrel one and a half times as much oil as he does the second barrel.
In math, we have $ x = 1.5y $, or in other words, $ y = \frac{2x}{3} $

The second condition is this:
He puts in the first barrel 15 gallons less oil in the third barrel. 
In math, we have $ x = z - 15 $, or in other words, $ z = x + 15 $

Of course, $ x + y + z = 167 $.

Putting this all together, we have


$ \begin{eqnarray*} x + y + z &=& 167 \\ x + \frac{2x}{3} + (x + 15) &=& 167 \\ 3x + 2x + 3(x + 15) &=& 501 \\ 3x + 2x + 3x + 45 &=& 501 \\ 8x + 45 &=& 501 \\ 8x &=& 456 \\ x &=& 57 \end{eqnarray*} $

Now we have $ 57 $ gallons of used oil in first barrel, $ \frac{2(57)}{3} = 38 $ gallons of used oil in the second barrel, and finally $ 57 + 15 = 72 $ gallons in the third barrel.

Cynthia

Problem:

Cynthia must cut a 113.4 inch piece of piece into three pieces. The first piece must be half as long as the second piece and 11.2 inches longer than the third. How long each piece should be.

Solution:

Let $ x $, $ y $ and $ z $ be the lengths of the first, second and third piece respectively.

We have this condition:

The first piece must be half as long as the second piece, that becomes

$ x = \frac{y}{2} $, or simply $ 2x = y $.

Next, we also have this condition:

The first piece must be 11.2 inches longer than the third, that becomes

$ x = 11.2 + z $, in other words, $ z = x - 11.2 $

Or course, the three pieces add up to 113.4 inches, that means

$ x + y + z = 113.4 $.

The simplest way to solve this is to represent all these in terms of  $ x $

$ \begin{eqnarray*} x + 2x + (x - 11.2) &=& 113.4 \\ 4x - 11.2 &=& 113.4 \\ 4x &=& 124.6 \\ x &=& 31.15 \end{eqnarray*} $

Therefore we get the the first piece is 31.15 inches, the second piece should be $ 31.15 \times 2 = 62.3 $ inches, the last piece is $ 31.15 - 11.2 = 19.95 $ inches.

Cost and revenue

Problem:



Solution:

Part a) f(x) = 10000 + 550x

This is because his total cost is the sum of the startup cost and the per computer cost

Part b) g(x) = 1199.99x

This is because he revenue is simply the revenue of selling the computers

Part c) g(14) - g(11)

This is the change in revenue.

Part d) f(c + d) - f(d)

This is the change in cost.

Part e) h(x)  = g(x) - f(x)

Profit is revenue minus cost

Part f) h(x) = 1199.99x - (10000 + 550x) = 649.99x - 10000

Part g) 16,

This is the minimal integer x that make h(x) > 0

Lookup

Problem:

Solution

a) No, we can see that clearly in this 'graph'






b) That would be -7.9, just read from the table

c) g(3.05) - g(-2) = 19.1 - (-5.6) = 24.7

d) Again, just do a backward lookup, we found t = 0

e) To find the rate of change, we compute

(g(-1.15) - g(-2)) / ((-1.15) - (-2)) = (1.9 - (-5.6)) / ((-1.15) - (-2)) = 8.823529412