Answer:
1,2,4,5,10
1,2,3,5,6
60
Step-by-step explanation:
Answer:
In algebra, an equation is a mathematical expression that contains an equals sign. ... An inequality is a mathematical expression that contains inequality signs. For example, y≤ 12x is an inequality. Inequalities are used to tell us that an expression is either larger or smaller than another expression.
Step-by-step explanation:
Answer:
the answer to this question is one solution
Answer: 160
Step-by-step explanation:1920 divided by 12
Answer:
or 
Step-by-step explanation:
Given
Points:
A(-3,2) and B(-2,3)
Required
Determine the radius of the circle
First, we have to determine the center of the circle;
Since the circle has its center on the x axis; the coordinates of the center is;

Next is to determine the value of x through the formula of radius;

Considering the given points



Substitute values for
in the above formula
We have:

Evaluate the brackets


Eva;uate all squares


Take square of both sides
Evaluate the brackets



Collect Like Terms


Divide both sides by 2

This implies the the center of the circle is

Substitute 0 for x

Substitute 0 for x and y in any of the radius formula


Considering that we used x1 and y1;
In this case we have that; 
Substitute -3 for x1 and 2 for y1


---<em>Approximated</em>