Problem:
Factorize 6x2−7y2
Solution:
(√6x+√7y)(√6x−√7y)
Friday, October 30, 2015
Thursday, October 29, 2015
Logarithm
Problem:

Solution:
6log(25−x2)−(log(5+x)+log(5−x))=6log(25−x2)−(log(5+x)(5−x))=6log(25−x2)−log(25−x2)=5log(25−x2)=log(25−x2)5
Angles
Problem:
Solution:
1) 180
2) PQS
3) 130
4) 30
5) 35
6) 100
7) 50
8) 130
9) x = 11, m<DEF = 29, m<FEG = 61
10) x = 10, m <DEF = 91, m<FEG = 89
11) 2, 4
12) 2, 3
13) Right
14) m<KLM = m<KLN = 45
Solution:
1) 180
2) PQS
3) 130
4) 30
5) 35
6) 100
7) 50
8) 130
9) x = 11, m<DEF = 29, m<FEG = 61
10) x = 10, m <DEF = 91, m<FEG = 89
11) 2, 4
12) 2, 3
13) Right
14) m<KLM = m<KLN = 45
Wednesday, October 28, 2015
Tuesday, October 27, 2015
Summation
Problem:

Solution:
ˉx=10.8.
Here are the values of absolute difference from the mean

Solution:
ˉx=10.8.
Here are the values of absolute difference from the mean
6.8 |
4.8 |
0.8 |
3.2 |
9.2 |
Here are the weighted absolute difference from the mean
27.2 |
14.4 |
1.6 |
3.2 |
9.2 |
The sum of these numbers is 55.6
Therefore the final result is 5.05454545...
Monday, October 26, 2015
Extra credits
Problem:

Solution:
f′(x)=3ax2+2bx+c
f″
The point of inflection has x coordinate -1, therefore
6a(-1) + 2b = 0
The extremas has x coordinate 1 and -3, therefore, 1 and -3 are the roots of 3ax^2 + 2bx + c = 0
3a + 2b + c = 0
3a(-3)^2 + 2b(-3) + c = 0 .
The three equations tell us b = 3a, c = -9a .
So the equation is f(x) = ax^3 + 3ax^2 - 9ax + d
When x = 1, f(x) = a + 3a - 9a + d = -5a + d = -9
When x = -1, f(x) = -a + 3a + 9a + d = 11a + d = 23
Therefore a = 2, d = 1
The final answer is 2x^3 + 6x^2 - 18x + 1

Solution:
f′(x)=3ax2+2bx+c
f″
The point of inflection has x coordinate -1, therefore
6a(-1) + 2b = 0
The extremas has x coordinate 1 and -3, therefore, 1 and -3 are the roots of 3ax^2 + 2bx + c = 0
3a + 2b + c = 0
3a(-3)^2 + 2b(-3) + c = 0 .
The three equations tell us b = 3a, c = -9a .
So the equation is f(x) = ax^3 + 3ax^2 - 9ax + d
When x = 1, f(x) = a + 3a - 9a + d = -5a + d = -9
When x = -1, f(x) = -a + 3a + 9a + d = 11a + d = 23
Therefore a = 2, d = 1
The final answer is 2x^3 + 6x^2 - 18x + 1
Point of inflection
Problem:

Solution:
The point of inflection is where the second derivative is 0, so we compute the second derivative as follow:
f'(x) = -3x^2 + 18x - 4
f''(x) = -6x + 18
Therefore the point of inflection has x coordinate 3, the y coordinate is -(3)^3 + 9 \times 3^2 - 4 \times 3 - 1 = 119
The curve is concave up in (-\infty, 3) and is concave down in (3, \infty) .

Solution:
The point of inflection is where the second derivative is 0, so we compute the second derivative as follow:
f'(x) = -3x^2 + 18x - 4
f''(x) = -6x + 18
Therefore the point of inflection has x coordinate 3, the y coordinate is -(3)^3 + 9 \times 3^2 - 4 \times 3 - 1 = 119
The curve is concave up in (-\infty, 3) and is concave down in (3, \infty) .
Sunday, October 25, 2015
Solid
Problem:

Solution:
Minimize 4\pi r^2 + 2\pi r h subject to \frac{4}{3}\pi r^3 + \pi r^2 h = 6 .
The Lagrangian is 4\pi r^2 + 2\pi r h - \lambda(\frac{4}{3}\pi r^3 + \pi r^2 h - 6) .
The partial derivatives must be 0, so we have
8\pi r + 2\pi h - \lambda (4 \pi r^2 + 2\pi r h) = 0
2 \pi r - \lambda(\pi r^2) = 0
The second equation implies \lambda = \frac{2}{r} .
Substitute this back to the first equation, solving get h = 0 .
This seems reasonable because we know sphere maximize volume and minimize surface area, it only make sense if the cylindrical part vanish.
So it is simple now, \frac{4}{3}\pi r^3 = 6 , r = \sqrt[3]{\frac{9}{2\pi}} = 1.127251652 .

Solution:
Minimize 4\pi r^2 + 2\pi r h subject to \frac{4}{3}\pi r^3 + \pi r^2 h = 6 .
The Lagrangian is 4\pi r^2 + 2\pi r h - \lambda(\frac{4}{3}\pi r^3 + \pi r^2 h - 6) .
The partial derivatives must be 0, so we have
8\pi r + 2\pi h - \lambda (4 \pi r^2 + 2\pi r h) = 0
2 \pi r - \lambda(\pi r^2) = 0
The second equation implies \lambda = \frac{2}{r} .
Substitute this back to the first equation, solving get h = 0 .
This seems reasonable because we know sphere maximize volume and minimize surface area, it only make sense if the cylindrical part vanish.
So it is simple now, \frac{4}{3}\pi r^3 = 6 , r = \sqrt[3]{\frac{9}{2\pi}} = 1.127251652 .
Farmer knew Lagrangian optimization!
Problem:

Solution:
Minimize x + 2y subject to xy = 405000
The Lagrangian is
L(x, y, \lambda) = x + 2y + \lambda(xy - 405000)
The partial derivatives must be 0, so we get
1 + \lambda y = 0
2 + \lambda x = 0
Multiply the first equation by x and multiply the second equation by y gives x - 2y = 0 .
In other words, x = 2y , so xy = 2y^2 = 405000 , y = 450, x = 900 .

Solution:
Minimize x + 2y subject to xy = 405000
The Lagrangian is
L(x, y, \lambda) = x + 2y + \lambda(xy - 405000)
The partial derivatives must be 0, so we get
1 + \lambda y = 0
2 + \lambda x = 0
Multiply the first equation by x and multiply the second equation by y gives x - 2y = 0 .
In other words, x = 2y , so xy = 2y^2 = 405000 , y = 450, x = 900 .
Friday, October 23, 2015
Limit
Problem:

Solution:
\frac{x^n - a^n}{x-a} = \frac{(x - a)(x^{n - 1} + ax^{n-2} + .. + a^{n-1})}{x - a} = (x^{n - 1} + ax^{n-2} + .. + a^{n-1})
Therefore \lim\limits_{x \to a}\frac{x^n - a^n}{x-a} = \lim\limits_{x \to a} (x^{n - 1} + ax^{n-2} + .. + a^{n-1}) = na^{n-1} .

Solution:
\frac{x^n - a^n}{x-a} = \frac{(x - a)(x^{n - 1} + ax^{n-2} + .. + a^{n-1})}{x - a} = (x^{n - 1} + ax^{n-2} + .. + a^{n-1})
Therefore \lim\limits_{x \to a}\frac{x^n - a^n}{x-a} = \lim\limits_{x \to a} (x^{n - 1} + ax^{n-2} + .. + a^{n-1}) = na^{n-1} .
Thursday, October 22, 2015
Finance 5
Problem:
Solution:
a)
Solution:
a)
Times
|
APY (Decimal)
|
APY (Percentage)
|
1
|
0.052
|
5.2%
|
12
|
0.053257411
|
5.325741057%
|
365
|
0.053371841
|
5.337184107%
|
Continuously
|
0.053375743
|
5.337574251%
|
b) 26912.4424
Finance 4
Problem:

Solution:
a) 2347.021742
b) f(t) = 2000 \times e^{\frac{3.2}{100}}
c) e^{\frac{3.2}{100}} - 1 = 0.032517505 = 3.251750531\%
d) (e^{\frac{3.2}{100}})^{10} - 1 = 0.377127764 = 37.71277643 \%
Solution:
a) 2347.021742
b) f(t) = 2000 \times e^{\frac{3.2}{100}}
c) e^{\frac{3.2}{100}} - 1 = 0.032517505 = 3.251750531\%
d) (e^{\frac{3.2}{100}})^{10} - 1 = 0.377127764 = 37.71277643 \%
Finance 2
Problem:

Solution:
To be continued with part b when the choices are available.
Solution:
Year
|
Monthly
|
Daily
|
Continuously
|
1
|
4825.305364
|
4826.254424
|
4826.286816
|
2
|
5174.127079
|
5176.162615
|
5176.232095
|
4
|
5949.24245
|
5953.924315
|
5954.084156
|
10
|
9043.476195
|
9061.279018
|
9061.887184
|
30
|
36524.23864
|
36740.36641
|
36747.76461
|
APY
|
7.229008086%
|
7.250098317%
|
7.250818125%
|
To be continued with part b when the choices are available.
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