1) x^2=36
x=6
3) x^2-8x+13=0 —> x= (8±√64-4(1)(13))/2(1)
x=4±√3
5) x^2-6x+9-k=0 —> x=(6±√36-4(1)(9-k))/2(1) —> (6±√4k)/2 —> (6±2√k)/2
x=3±√k
7) y=x^2-4x+11 —> y-11=x^2-4x —> take the half of the coefficient of the single x term and square it and add it on both sides —> y-11+4=x^2-4x+4 —> y-7=(x-2)^2 —> y=(x-2)^2+7
Minimum: (2,7)
Maximum: n/a
X intercepts: none (never crosses the x-intercept)
9) y=x^2+2x-8 —> y+8+1=x^2+2x+1 —> y=(x+1)^2-9
Minimum: (-1,-9)
Maximum: n/a
x-intercepts: (x+4)(x-2) —> (-4,0),(2,0)
11) c
13) (x+7)(x+3)
15) x=(-6±√36-4(1)(10))/2 —> x=(-6±√-4)/2 —> (-6±2i)/2
x=-3±i OR no real solutions
We can use tangent to calculate the length of ST.
Here, the opposite side of the 60 degree angle is ST, and the adjacent is RS. Plug in what we know, we can use 'x' for side ST:
Solve for 'x', multiply 7 to both sides of the equation:
This is an exact answer for the length of ST, we can find an approximate with a calculator:
You cannot add row 2 to column 3 because they have different dimensions. You can do any of the other operations, but the only one that makes any sort of sense is ...
Multiply row 2 by -1 and add it to row 3
_____
It makes no sense to multiply a row by zero. That makes the entire row zero and makes the matrix useless for finding any sort of solution.
You can switch columns, but that doesn't get you any closer to a solution here.
If I were trying to find a solution, I might
switch rows 1 and 2
multiply the new row 1 by -3 and add it to the new row 2
multiply the new row 1 by 2 and add it to row 3
This sequence of operations will make the first column [1 0 0], reducing the problem to 2×2 from 3×3.
8/6 = 12/9
8/6 x 9/9 = 72/54
12/9 x 6/6 = 72/54
72/54 = 72/54
They are similar.
Triangle ABC is similar to triangle PQR
