## Archive for the ‘Geometry’ Category

## Problem 2043

The slope of a given line is shown below. Find the slope of a line parallel to the given line and the slope of a line perpendicular to the given line.

m = 3

**Solution:-**

The slope of a line parallel to the given line is m = 3

The slope of a line perpendicular to the given line is m =

## Problem 2042

The slope of a given line is shown below. Find the slope of a line parallel to the given line and the slope of a line perpendicular to the given line.

m = -12

**Solution:-**

First find the slop of a line parallel to the given line.

Two nonvertical lines are parallel of and only of their slope are equal and they have different y-intercepts. Vertical lines are parallel if they have different x-intercepts.

Therefore, the slope of a line parallel to the given line is m = -12.

Now find the slope of a line perpendicular to the given line.

Two nonvertical lines are perpendicular if and only if the product of their slopes is -1. Put another way, two nonvertical lines are perpendicular if their slopes are negative reciprocals of each other. Any vertical line is perpendicular to any horizontal line.

To find the slop of a line perpendicular to a given line, determine the negative reciprocal of the slope of the given line.

The negative reciprocal of -12 is or .

Therefore, the slope of a line perpendicular to the given line is

m =

## Problem 2037

Explain why there are restrictions on the domain for the function

f(x) =

**Solution:-**

The domain of a function is the set of values for which it is defined. For the given function *x* = 0 will cause the function to be undefined, since 1 divided by 0 is undefined. Therefore, the domain is the set of real numbers, except x = o or x ≠ 0.

## Problem 2030

Find the supplementary angle of 122.3◦

**Solution:-**

The supplementary angle of 122.3◦ is 57.7◦

## Problem 2029

Find the measure of the complement of a 52◦ angle.

**Solution:-**

The measure of the complement of a 52◦ angle is 38◦.

## Problem 2028

Identify the figure as a line, half line, ray, line segment, open line segment, or half open line segment. Then, name the figure using the given points.

Choose the correct description below.

Choose the correct symbol below.

**Solution:-**

Choose the correct description below.

Open line segment

Choose the correct symbol below.

## Problem 1112

What point is symmetric with respect to the y-axis to the point ?

**Solution:-**

Recall for any point (x, y) the point symmetric about the y-axis to (x, y) is the point (-x , y).

If two point are symmetric about the y-axis, then the points have the same y-coordinate and opposite x-coordinates. Replace the x-coordinate of the given point with its opposite.

Identify the x-coordinate of the given point, .

x =

If x = ,then the opposite of x, -x = –

Therefore , the point symmetric to the point with respect to the y-axis is .

## Problem 1035

Solve the inequality symbolically. Express the solution set in interval notation.

-7(z-8)≥4(9-2z)

**Solution:-**

To solve the inequality, isolate the variable on one side the inequality using the properties if inequalities.

First, use the distributive property to eliminate the parentheses on both sides of the equation.

-7(z-8) ≥ 4(9-2z)

-7z + 56 ≥ 36 – 8z (Simplify the right side.)

Add 8z to both sides so that all expressions containing a variable are on the left side.

-7z +56 ≥ 36 – 8z

z + 56 ≥ 36 (Add 8z to both sides)

To isolate z, subtract 56 form both sides.

z + 56 ≥ 36

z ≥ 56 ≥ 36

z ≥ -20 (Subtract)

Thus, the solution is the set of all real numbers greater than or equal to -20.

In interval notation, a square bracket is used to show that a number is part of the interval and a parenthesis is used to indicate that a number is not part of the interval.

Write the solution set is interval notation.

[-20,∞)

## Problem 986

**Polynomial conditions.**

a) If the degree is even and the leading coefficient is positive, the polynomial rises on the left and rises on the right.

b) If the degree is even and the leading coefficient is negative, the polynomial falls on the left and falls on the right.

c) If the degree is odd and the leading coefficient is positive, the polynomial falls on the left and rises on the right.

d) If the degree is odd and the leading coefficient is negative, the polynomial rises on the left and fall on the right.