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:
Step-by-step explanation:
it's obtuse the sides are to wide
Answer:
Step-by-step explanation:
∠ABC = ∠BCD { AB // CD; BC TRAVERSAL, alternate angles are equal}
∠ABC = 20
∠BCA = ∠BAC {angles opposite to equal sides are equal}
∠BCA = ∠BAC = x
x + x + 20 = 180
2x = 180 - 20
2x = 160
x = 160/2
x = 80
Answer:
Step-by-step explanation:
44 sq inches, i am so not sure.
