﻿ Assignment~ | Researchomatic

Assignment~

 Tweet

ASSIGNMENT~

Assignment~

Part A

Solve the following for x

Solution

The procedure used to solve this question is called substitution. In this method, the complicated term is replaced by a variable. The algebraic methods are than used to solve the new form of the equation for the new variable. When the values are obtained, the new variable is then replaced by the expression of the original variable, and again soved for the original expression.

Let t = vx+1

The equation becomes

4/t - 3t +1 = 0

Multiplying both sides by t

4 - 3t2 + t = 0

Or 3t2 - t - 4 = 0

3t2 - 4t + 3t - 4 = 0

t(3t - 4) + 1(3t - 4) = 0

(t+1)(3t-4) = 0

Either t+1 = 0

t = -1

or 3t+4 = 0

t = -4/3

substituting the expression of x in place of t

vx+1 = -1

Squaring both sides

x+1 = 1

x= 0

also

vx+1 = -4/3

Squaring both sides

x+1 = 16/9

x = (16/9) -1

x = (16-9)/9

x = 5/9

Thus, solution set of x is 0 and 5/9.

Determine the value of a if is a real number, and find this number.

Solution

The above question is of complex numbers. Complex numbers comprise of real and imaginary numbers. Imaginary numbers are those that do not exist on a real line. In the given expression, i is an imaginary number that is equal to v(-1).

Rationalizing the above expression, through multiplying and dividing by 1+v7 i.

The expression thus become,

= (v7 + ai)(1 + v7 i) / (1 - v7 i) (1 + v7 i)

= (v7 + ai)(1 + v7 i) / (1 - (v7 i)2 )

= (v7 + ai)(1 + v7 i) / (1 - 7(-1))

= (v7 + ai)(1 + v7 i) / 8

= v7 / 8

Solve the following inequalities

Solution

Inequalities are solved same as equalities except the answer is a range instead of a unique or multiple numbers. We can add or subtract numbers on both sides of an inequality. If a number is multiplied or divided on both sides of the inequality, the inequality sign reverse.

We know that (x2 - 9) = (x + 3)(x - 3) as 9 is the square of 3.

Substituting this

x2 - 3x +2 = x - 3

x2 - 3x + 2 - x + 3 = 0

x2 - 4x + 5 = 0

Find the equation of the tangent at the point (-2, 2)on the circle

State the value of the shortest distance from the center of the circle to the tangent.

Solution

The given point is x1 = -2 and y1 = 2

x2 + y 2 + 4x - 4y - 17 = 0

where A = 4 and B = -4

the formula to calculate the slope of the line tangent to the given circle at point (x1, y1) is given by

m = (2x1 + A)/(2y1 + B)

m = {2(-2) + (-4)} / (2(2) + (-4)}

m = -8/0

m = infinity

Thus the tangent is y = 2.

The shortest distance from the centre to the tangent of the circle is the radius of the circle.

For radius completing square for x and y

x2 + 4x + ...
• Agency Visit Assignment
www.researchomatic.com...

Agency Visit Assignment , Agency Visit Assi ...

• Math Assignment
www.researchomatic.com...

Math Assignment , Math Assignment Essay ...

• Case Assignment And Sessi...
www.researchomatic.com...

Case Assignment and Session Long Project Case ...

• E-Commerce Assignment
www.researchomatic.com...

E-Commerce Assignment , E-Commerce Assignme ...

• Mis Assignment For Mba
www.researchomatic.com...

Mis Assignment For Mba, Mis Assignment ...