Intermediate algebra test

Intermediate algebra is mainly used for solving the equations. Intermediate algebra mainly uses the simplification technique for solving the equations. Intermediate algebra uses the letters and variables. Intermediate algebra also deals with the expressions and exponents of polynomials. Algebra test problems are very simple and easy. Intermediate algebra test covers the topics,

• linear Equations
• Equation of a line
• Quadratic Equations
• Polynomial Functions
• Slope formula

Intermediate Algebra Test Problems

Algebra test Problem: 1

Solve the equation x + y = 6, x - 3y = 2.

A) x = 2, y = 4 B) x = 5, y = 2 C) x = 4, y = 2 D) x = 5, y = 1.

Answer: D

Algebra test Problem: 2

Find the slope of the equation 4x + 3y = 21

A) M = - 4 / 5 B) M = - 4 / 3 C) M = - 3 / 4 D) M = - 3 / 5.

Answer: B

Algebra test Problem: 3

Factorize the equation 2x2 - 5x + 3.

A) X = 2, X = 4 B) X = - 2, X = - 3 C) X = 1, X = 3 / 2 D) X = 1, X = 2 / 3.

Answer: C

Algebra test Problem: 4

Find the vertex of the equation Y = x2 + 4x + 7

A) (2, 4) B) ( -2, 3) C) (-3, 2) D) (1,-1).

Answer: B

Algebra test Problem: 5

Find the distance between the two points (1, 5) and (3, 9).

A) d = 6 B) d = 3 C) d = 7 D) d = 5.

Answer: D

Problems for Intermediate Algebra Test

Algebra test Problem: 6

 Find the value of x for the expression: 4x + (4 - 7x) = (2x + 3) - 4.

A) 3 B) 2 C) 8 D) 1

Answer: D

Algebra test Problem: 7

Add the two expressions (3x + 2y - 8) and (5x - 3y -4)

A) 5x + 4y - 12 B) 8x + y + 12 C) 8x - y - 12 D) 2x - y - 4.

Answer: C

Algebra test Problem: 8

The sum of the number and 32 is 67. Find the unknown number.

A) 32 B) 33 C) 31 D) 35

Answer: D

Algebra test Problem: 9

Multiply (3a2 + 4b) (2a + 6b)

A) 6a3 + 8ab + 18a2b + 24b2

B) 3a3 + 8ab + 18a2b + 24b2

C) 4a3 + 6ab + 16a2b + 24b2

D) 12a3 + 9ab +18a2b + 24b2

Answer: A

Solving Trigonometric Equations made easy

The six trigonometric functions are the sine, cos, tan, csc, sec and cot. An equation consisting of one or more trigonometric functions is called a trigonometric equation, for instance cot^2(x) + 1 = csc^2(x) is a trigonometric equation.  Some of Trigonometric Equations Examples are 2 cos^2(x) –sin(x) – 1 =0, 2sin(theta) – 1 = 0, cos(theta/2) = 1 + cos(theta), 6 sin^2(theta) – sin(theta) = 1, cot(2 theta) – tan(2 theta) = 0

Quadratic Trigonometric Equations
Trigonometric equations that involve trigonometric functions as products are called the quadratic trigonometric functions. In a quadratic equation with degree 2 written in the form of  ax^2 + bx + c. So, a quadratic trigonometric equation will consist of trigonometric functions with degree 2. For instance, 2 sin^2(x) – sin(x) +1 = 0 is an example for a quadratic trigonometric equation.

Let us now learn the steps involved in simplifying trigonometric equations. We solve trigonometric equations as we solve any other algebraic equation except that we use the trigonometric identities in solving. They are also solved by the sign of the trigonometric value by determining the quadrant(s) for that particular trigonometric value. Let us simplify a trigonometric equation, 2 sin(pi/2 -alpha) =1. We shall use the identity sin (pi/2 – alpha) = cos(alpha) in solving the given equation. So, the trigonometric equation reduces to, cos(alpha)=1/2 . Now we need to check for what angle of cosine alpha gives the value ½. We know that cos(pi/3) equals ½. So, we write it as cos(+/-) pi/3 =1/2. The positive and negative sign depends  on which quadrant the angle alpha lies and hence we can generalize the solution as alpha = (+/-)pi/3 + 2k pi where k is any integer. There is a significance for adding 2k pi to the solution, as per the periodic behavior of the function cosine. So, the solution set is written as {alpha = (+/-)pi/3 + 2k pi: k belongs to set of integers}

Sin(x) + cos(x) = 0
Simplifying the above equation we get, sin(x) = -cos(x). Here, the value of sine and cosine have to be equal with opposite sign. Let us write sine in terms of cosine using trigonometric identity, which gives us
  cos(pi/2 – x+2 k pi) = cos(pi + x)
      pi/2 – x + 2 k pi = pi + x

The general solution would be, x = -pi/4 + k pi where k belongs to set of integers
Now, in the interval [0, 2pi) let us consider k = 1 and k = 2, we get, x = (3/4) pi and x = (7/4) pi as the solution that is at 135 degrees and 315 degrees the value of sine and cosine are equal and opposite.

Steps for solving standard deviation problems

A standard deviation example would typically have a set of observations x_1, x_2, x_3,  …… x_n. The number of observations being n. In general it can be written like this: x_i, where i = 1, 2, 3, … n. To calculate the standard deviation of such a problem the following steps are followed:

Step 1: List all the observations in the first column of a table. The title of column would be x_i and the observations would be x_1, x_2, x_3,  …… x_n.

Step 2: Find the mean of the observations. If we denote the mean by x ¯, then the formula for calculating the sample mean would be:  x ¯ = (x_1+ x_2 + x_3+  ……+ x_n)/n = [∑x_i]/n

Step 3: We now come to the second column. In this column we find the deviation of each of the observation from the mean. For that we calculate the value of x_i - x ¯ for each of the observations.
Note that in this step it is possible that we get negative values. We shall take care of the negatives in the next step.

Step 4: Now we construct the third column. In this column we square the results of column (3) for each of the observations. That means we find the value of (x_i - x ¯)^2 for each of the n observations. Note here that since we squared the difference, the negatives are got rid of so we need not worry about the negative numbers canceling out the positive numbers and affecting our end result.

Step 5: Now we find the standard deviation for each of the observation. For that we compute the following value: (x_i - x ¯)^2 / n. That would be placed in our fourth column.

Step 6: Adding standard deviations for each of the observations we get the value of the expression : ∑((x_i - x ¯)^2 / n). That will give us our sample standard deviation formula. The symbol for standard deviation is the Greek small alphabet ‘s’ pronounced as ‘sigma’. Therefore we say that:
s = sqrt[∑((x_i - x ¯)^2 / n)] or s = sqrt [∑(x_i - x ¯)^2/n]

Inverse Functions and their derivatives in a nutshell

In Math, Inverse Function is a function whose relation to a given function is such that their composite is the identity function. Inverse Function can also be defined as the function obtained by expressing the dependent variable of one function as the independent variable of another; for instance f and g are inverse-functions if f(x)=y and f(y)=x. It is denoted by (-1) in the power, inverse of f(x) is f-1(x).

One important point we need to remember is that the inverse of a function may not always be a function. While finding Inverse of Function of a function y in terms of x, we just switch the x and y and then solve for y. The new function [y-1] we get is the inverse-function of the given original function. For instance, finding Inverse of a Function f(x)= (x/3) -1 involves a list of steps. First let us take f(x)=y=(x/3) -1. In the next step we switch the x and y, that gives us, x = (y/3)-1. The last step would be to solve for y, finally we get y= y-1= 3x+3 which is the inverse-function of f(x) = (x/3) -1

There are four main steps involved in Solving Inverse Functions. Given a function f(x) in terms of x,
1. First step, we write the given function f(x) equal to y
2. Second step involves interchanging the x and y
3. In the third step, we solve for y
4. Fourth step, we write y as f-1(x) which is the inverse-function of f(x)

Considering an example, let us now learn how to solve Inverse functions using the above steps. Given the function f(x) = 2x-3, let us solve inverse-function [f-1(x)]. The first step would be to write y = 2x-3. Next we interchange x and y which gives, x = 2y-3. Now, we solve for y; y = (x+3)/2. The last step is to put f-1(x) in place of y which gives us f-1(x)= (x+3)/2  which is the inverse-function of the given function.

Derivative of an Inverse Function, if for a function f an inverse-function f-1 exists and if f is differentiable at f-1(x) and f’[f-1(x)] is not equal to zero, then f-1 is differentiable at x and the formula is given as:

Considering an example, let us now learn how to solve Inverse functions using the above steps. Given the function f(x) = 2x-3, let us solve inverse-function [f-1(x)]. The first step would be to write y = 2x-3. Next we interchange x and y which gives, x = 2y-3. Now, we solve for y; y = (x+3)/2. The last step is to put f-1(x) in place of y which gives us f-1(x)= (x+3)/2  which is the inverse-function of the given function.

Derivative of an Inverse Function, if for a function f an inverse-function f-1 exists and if f is differentiable at f-1(x) and f’[f-1(x)] is not equal to zero, then f-1 is differentiable at x and the formula is given as:

A square root function is the inverse-function of a square function f(x) = x2. So, the derivative of an Inverse Function of square function f(x) = x2 can be solved using the power rule

An Introduction to Quartiles in Statistics

Quartile is a term derived from the word ‘quarter’ which is one fourth of something. So, we can say  Quartiles Statistics is a certain fourth of a data set. When a given data is arranged in the ascending order that is from the lowest value to the highest value and this data is divided into groups of four, we get what we call Statistics Quartiles.

In Quartiles Statistics there are three quartiles which are, the first quartile also called as lower quartile is denoted as Q1 which lies in the twenty five percent of the bottom data. The second quartile is the median of the data that divides the data in the middle and has fifty percent of the data below it and the other fifty percent of the data above it. It is denoted as Q2.The third quartile also called the upper quartile denoted as Q3 has seventy five percent of the data below it and twenty five percent of the data above it.

We can state the Quartiles Definition as, Quartiles are the three values that divide an ordered data set into four approximately equal parts. Definition of Quartiles can also be given as, A statistical term describing a division of a data set into four defined intervals based upon the values of the data and how they compare the entire data set.

Now that we are a bit familiar with Quartiles Statistics, let us learn how to calculate Quartiles of a given data set. For a better understanding of the steps involved let us consider a data set. Calculate the three
 quartiles of the data set of scores,  27 18 20 20 23 29 24.

The steps involved are:
• Count the total number of scores in the given data; n= 7
• Arrange the data set of scores in the ascending order, that is, from the lowest score to the highest score, we get, 18 20, 20, 23, 24, 27, 29
• Find the median (Q2) which is the middle value of the data set
Median = ½(n+1) = ½(7+1)= 4
So, the median (Q2)=23 [the fourth term of the ordered data set]
• Next we shall calculate the lower quartile (Q1) which is the median of the lower half of the data set.    Q1 = ¼(n+1) = ¼(7+1) = 2
So, the lower quartile (Q1) = 20 [the second term of the ordered data set]
• Finally we shall calculate the upper quartile (Q3) which is the median of the upper half of the data set. Q3=3/4(n+1) = ¾(7+1) = 6
So, the upper quartile (Q3)= 27 [the sixth term of the ordered data set]

Please express your views of this topic Statistics Problems by commenting on blog.

Derivatives of Cot functions for some special values

We can find the derivative of the cot function using the quotient rule of differentiation. The quotient rule can be states as follows: For a function f such that f(x) = u/v, where u and v are both function of x, the derivative of f(x) is given by the formula:
f’(x) = (v*u’ – u*v’)/v^2, where u’ is derivative of function u and v’ is derivative of function v.
We know that, cot x = cos x / sin x,
Therefore, (d/dx) cot x = (d/dx) (cos x / sin x),
= [sin x (d/dx) (cos x) – cos x (d/dx) (sin x)]/sin^2 (x)  … using the quotient rule stated above.
=[ sin x (-sin x) – cos x * cos x]/sin^2(x)
= [-sin^2(x) – cos^2(x)]/sin^2(x)
= -[sin^2(x) + cos^2(x)]/sin^2(x)
= -1/sin^2(x)
= -csc^2(x)
Therefore the derivative of cot x is –csc^2(x)

The above derivative can also be proved using the definition of derivative which is like this:
F’(x) = lim(t->x) [(f(t) – f(x)]/(t-x), for the function y = f(x) = cot (x) the definition would be like this:
(d/dx) cot x = lim(t->x) [cot t – cot x]/(t-x)
= lim(t->x) [1/tan t – 1/tan x]/(t-x)
  = lim(t->x) [tan x – tan t]/[(tan t * tan x)(t-x)]
= lim(t->x) {[tan x – tan t]/(t-x)} * 1/[tan t * tan x]
= lim(t->x) {-[tan t – tan x]/(t-x)} * lim(t->x) 1/[tan t * tan x]
= lim(t->x) {-[ tan(t-x) (1+tan t tan x)]/(t-x)} * 1/tan^2(x)
We know that lim(t->x) of tan(t-x)/(t-x) = 1. Therefore,
= -{1+tan^2(x)} *1/tan^2(x)
= - sec^2(x)/tan^2(x)
= [-1/cos^2(x)]/[sin^2(x)/cos^2(x)] … here the cos^2(x) cancel each other out.
= -1/sin^2(x)
= -csc^2(x)

Derivative of cot-x:
For finding the derivative of cot(-x) we use the chain rule.
Let –x = u, then du/dx = -1.
Cot (-x) = cot u
(d/du) cot u = - csc^2(u) …  from what we proved above.
Using the chain rule we have:
(d/dx) cot (-x) = (d/dx) cot u * du/dx = -csc^2(u) * (-1)
= -csc^2(-x) * (-1)
= csc^2(-x)
Thus we see that the derivative of cot(-x) is csc^2(-x)

Derivative of cot^-x (k):
Cot^(-x) (k) where k is some constant can be written like this also
= (cot(k))^(-x)
= 1/(cot k)^x
= (1/cot k)^x
Since k is a constant, cot k is also a constant and in turn, 1/cot k is also constant. So let 1/cot k = some constant say p. Therefore,
= p^x is the function.
(d/dx) p^x = p^x ln p
= (1/cot k)^x ln(1/cot k)
That is the derivative of the above function.

Triple integrals spherical co-ordinates

The concept of definite integration can be extended to two or three or multi dimensional system. For a bounded function f(x,y,z) defined on a rectangular box B (x0 = x = x1, y0 = y = y1, z0 = z = z1), the triple integral of f over B,
Integral (B) [f(x,y,z)] dV or  Integral  (B) [f(x,y,z)] dx dy dz,
can be defined as a suitable limit of Riemann sums corresponding to partitions of B into sub boxes by planes parallel to each of the co-ordinate planes. Triple integrals over more general domains are defined by extending the functions to be zero outside the domain and integrating over a rectangular box containing the domain.
All properties of double integrals hold good for triple integrals as well. In particular, a continuous function is integrable over a closed, bounded domain. If f(x,y,z) = 1 on the domain D, then the triple integral gives the volume of D.
Volume of D =  Integral (B)  dV
Triple integrals spherical coordinates: The spherical co-ordinates related to Cartesian co-ordinates x, y and z are given by the equations:

The triple integrals in spherical coordinates relationships are illustrated in the picture below

For the co-ordinate surfaces in the spherical co-ordinates, see picture below:

The volume in spherical co ordinates would be like this:

Spherical co ordinates are suited to problems involving  spherical symmetry and, in particular, to regions bounded by spheres centered at origin, circular cones with axes along the z axis, and vertical plains containing the z axis.