represents the set that contains all the elements that are in at least one of the sets, so 
.
We want the complement of this set (in other words, the set with all the elements in the universal set but not in the given set).
This set is {3, 4, 11, 13, 14, 15, 17, 24}
Remark
This is a very interesting question. Draw a line from the origin to where the upper right vertex of the square touches the line. That line has the property that the its equation is y = x. So the "solution" to the point of intersection is the solution of the two equations.
y = x (1)
3x + 4y = 12 (2)
Put x in for y in equation 2
3x + 4x = 12
7x = 12
x = 12/7
x = 1.714
y = 1.714
Problem A
<em><u>x intercept</u></em>
The x intercept occurs when y = 0
3x + 4(0) = 12
3x = 12 Divide by 3
x = 12/3
x = 4
the x intercept = (4,0)
<em><u>y intercept</u></em>
The y intercept occurs when x =0
3(0) + 4y = 12
4y = 12
y = 12/4
y = 3
y intercept = (0,3)
Problem B
x and y both equal 1.714 so they are also the length of the square's side.
Problem C
See solution above. x =y is the key fact.
x = y = 1.714
Answer:
22
Step-by-step explanation:
BD bisects <ABC, this means the half-line divided the angle into two equal parts.
If m<ABC is equal to 44 then m<ABD is
44/2 = 22
Answer:
49π
Step-by-step explanation:
πr^2
π*7^2
49π
