Thursday, September 27, 2012

Exponents Negative Numbers

Introduction:

         The exponent number is shows how many times to use the base number in a multiplication. The exponent number is placed at upper right of the base number. Exponent -x in the expression a-x. For example, -4 is the exponent in 2-4= 1/24 = 0.0625. The base and exponent numbers may be positive or negative. Negative exponents are a way of indicating reciprocals.

Rules of Exponents Negative Numbers

Definitions
1. an = a·a·a···a  (n times)
2. a0 = 1  (a ≠ 0)
3. a-1 = 1/an (a  ≠ 0)
4. am/n = n√am or (n√a)m (a ≥ 0, m ≥ 0, n > 0)
Combining
1. Multiplication: ax ay = ax + y
2. Division: ax / ay = ax-y (a ≠ 0)
3. Powers: (ax)y = axy
Distributing   (a ≥ 0, b ≥ 0)
1. (ab)x = ax bx
2. (a/b)x = ax/bx  (b ≠ 0)
Careful!!
1. (a + b)n ≠ an + bn
2. (a – b)n ≠ an – bn
Rule for Exponents Negative Numbers:
a-n = 1/an
Examples: 5-2 = 1/52 = 1/25 
(2/3)-3 = (3/2)3 = 27/8

Exponents Negative Numbers – Examples
Negative exponents numbers solved problems
Example 1: Solve this expression 8-2
Solution:
     Here, the exponent is -2 negative exponents. Usually in positive exponent the exponent number is shows how many times to use the base number in a multiplication. In negative exponent number also shows like this. But in the negative exponent numbers we have to find the reciprocal of the numbers.
8-2 = 1/82 = 1 / (8 × 8) = 1/64 = 0.015625
82 = 8 × 8 = 64.
Reciprocal of the number is 1/64 = 0.015625
Example 2: Solve 4-3
Solution:
4-3 = 1/43 = 1 / (4 × 4 × 4) = 1/64 = 0.015625
Example 3: Solve this expression 22/2-3
Solution:
22/2-3 = 22 × 23 = 25 = 2 × 2 × 2 ×2 × 2 = 32
Example 4: Solve this expression 2(3-1)
Solution:
2(3-1) = 2(1/3) = 2/3 = 0.67
Example 5: Simplify this equation and solve this equation (4x)-3, x = 2.
Solution:
(4x)-3 = 1/64x3
Put x = 2
1/64(23) = 1/64(8) = 1/512 = 0.001953125
Example 6: Simplify this equation and solve this equation (x-3/y-4)-3, x = 1, y = 1.
Solution:
(x-3 / y-4)-3 = (x-3)-3/(y-4)-3 = (y-4)3/(x-3)3 = y-12/x-9 = x9/y12
Put x =1 and y = 1 in the equation to get
19/112 = 1/1 =1


Exponents Negative Numbers – Practice

Solve these problems for practice on negative exponents.
Problem 1: Solve 2-2 - Answer: 0.25
Problem 2: Solve 3-2 - Answer: 0.11
Problem 3: Solve 32/3-3 - Answer: 243
Problem 4: Solve this expression 4(4-1) - Answer: 1
Problem 5: Simplify this equation and solve this equation (2x)-2, x = 3. - Answer: 36

Monday, September 24, 2012

Precalculus Calculator Online

Introduction :

Precalculus calculator online is one interesting topics in mathematics. Precalculus calculator online is used to solve different types of precalculus problems. Calculator is a web-based tool to solve the problems. Online is nothing but the one computer is connected with another computer through a network or a cable. Here we solve some precalculus calculator online problems.

Example Problems for Online Precalculas Calculator:

Example problems for online precalculas calculator are given below:

Example 1:

Solve the quadratic equation x2 + x – 42.

Solution:

Let f(x) = x2 + x – 42

Now, plug f(x) = 0

x2 - 6x +7x - 42 = 0

x(x - 6) + 7(x - 6) = 0

(x - 6)(x + 7) = 0

x = 6; x = -7

The roots are x = 6, x = -7.

Example 2:

Solve 12x – 4y + 20 = 0. Find the slope and y-intercept for the given straight line.

Solution:

12x – 4y + 20 = 0

– 4y = – 12x – 20

Dividing by -4,

y = 3x + 5 ? (1)

General form of a straight line is,

y = mx + b ? (2)

Where, m = slope of a line,

b = y intercept of a line,

Here, y = 3x + 5

Compare the equation (1) and (2), we get,

Slope of the line m = 3,

y-intercept of the line b = 5.


Additional Example problems for online precalculas calculator are given below:

Example 3:

Find the center and radius of the circle for the given standard equation x2 + 10x + y2 – 8y – 7 = 0

Solution:

Given: x2 + 10x + y2 – 8y – 7 = 0

Standard equation for circle with center (a, b) and radius r is,

(x - a)2 + (y - b)2 = r2

Completing the x terms and y terms on the square that gives

(x2 + 10x + 10) + (y2 - 8y + 8) – 7 - 10 - 8 = 0

(x2 + 10x +10) + (y2 - 8y + 8) = 7 + 10 + 8

(x + 10)2 + (y - 8)2 = 25,

Solution to the center of the circle is (10, -8), and the radius is 5.

Example 4:

Find the vertex of the parabola y = 5x2 – 30x + 9

Solution:

General form:

x-coordinate for the vertex of the parabola is x = -b/2a,

y-coordinate is find by substitute the value for x into f(x)

Given:  y = 5x2 – 30x + 9

We know that x = -b/2a,

Here a = 5, b = -30

So that,   X = -b/2a = -(-30)/(2*5) = 3

And then y = 5(32) – 30(3) + 9 = 45 – 90 + 9 = -36

Solution to the problem is x = 3 and y = -36.