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:
1 root
Step-by-step explanation:
A linear equation should only cross the x-axis at one point. That means there would only be 1 root. (Root is when y= 0)
Your answer would be a because of the state of the appropriate bill and alternate hypotheses makes the most sense out of B and C
Answer:
18 and 19
Step-by-step explanation:
Answer:
The lower quartile of Class A is greater than the lower quartile of Class B.
Class A has a Greater number of lower quartiles then Class B as the imagie shows. Class A has 65-75 while class B has 70-75.
