The original number is 7 i think.
10(x^2) = 70x
10x^2 - 70x = 0
x^2 -7x =0
Δ= b^2 -4ac
where c doesnt exist=0
Δ= 49
-b ± √Δ /2a
7 ± √49 /2
7±7/2
x is either 14/2= 7
or 1/2, which doesnt work.
Lets see:
10(7^2) = 70•7
10•49 = 490
and that is right.
i hope my logic is correct.
8.

9.
The sum of two sides must be greater than the third side.
It can't be 5 because

.
It can't be 7 because

.
It can't be 17 because

.
So it's 12.
10.
a. Turk, McAllister
b. Market, Mission, Howard
c. not sure about this one
Some points are (7,3) and (-7,-5).
cot(<em>θ</em>) = cos(<em>θ</em>)/sin(<em>θ</em>)
So if both cot(<em>θ</em>) and cos(<em>θ</em>) are negative, that means sin(<em>θ</em>) must be positive.
Recall that
cot²(<em>θ</em>) + 1 = csc²(<em>θ</em>) = 1/sin²(<em>θ</em>)
so that
sin²(<em>θ</em>) = 1/(cot²(<em>θ</em>) + 1)
sin(<em>θ</em>) = 1 / √(cot²(<em>θ</em>) + 1)
Plug in cot(<em>θ</em>) = -2 and solve for sin(<em>θ</em>) :
sin(<em>θ</em>) = 1 / √((-2)² + 1)
sin(<em>θ</em>) = 1/√(5)
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