The error to that problem is the person switched the x and y points, like in point A, it’s supposed to be (x,y) or (1,5). The person put the y in front of the x while writing the points.
Answer:
xy = 1
k = 79
Step-by-step explanation:
Question One
The first and third frames look to me to be the same. I'll treat them that way.
y = x^2 Equate y = x^2 to the result of 2y + 6 = 2x + 6
2y + 6 = 2(x + 3) Remove the brackets
2y + 6 = 2x + 6 Subtract 6 from both sides
2y = 2x Divide by 2
y = x
Now solve these two equations.
so x^2 = x
x > 0
1 solution is x = 0 from which y = 0. This won't work. x must be greater than 0. So the other is
x(x) = x Divide both sides by x
x = 1
y = x^2 Put x = 1 into x^2
y = 1^2 Solve
y = 1
The second solution is
(1,1)
xy = 1*1
xy = 1
Answer: A
Question Two
square root(k + 2) - x = 0
Subtract x from both sides
sqrt(k + 2) = x Square both sides
k + 2 = x^2 Let x = 9
k + 2 = 9^2 Square 9
k + 2 = 81
k = 81 - 2
k = 79
Answer:
Step-by-step explanation:
In ΔBCA and ΔDCA,
BC ≅ CD Given
∠BCA ≅∠DCA Given
AC ≅ AC Reflexive property
ΔBCA ≅ ΔDCA S A S congruent
∵ BA ≅ AD CPCT {Corresponding part of congruent triangle}
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