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
Step-by-step explanation:
2x+1x= -2+1
3x= -1
3x÷3= -1÷3
x=|-3.3333..|=3.33333.. or
64 and 28 have a GCF of 4, since 64 = 4 x 16, 28 = 4 x 7, and 7 and 16 have no factors other than 1 in common. Knowing that, we can rewrite 64 + 28 as
4 x 16 + 4 x 7
and then use the distributive property to rewrite it again as
4 x (16 + 7)
