<span>t^2-4t-32= 0, implies sqrtDelta = 6
so t = 2 - 6= -4, or t = 2+6= 8
so </span>t^2-4t-32/t-8 = (<span>t + 4) (t-8) / t-8 = t+4, it is the simplified form
so the answer is </span><span>c. t + 4; t = 8</span>
Answer:
To solve this problem, we need to use the following two facts:
1) If a quadratic equation has integer coefficients only, and if one of the roots is a + √b (where a and b are integers), then a - √b is also a root of the equation.
2) If r and s are roots of a quadratic equation, then the equation is of the form x^2 – (r +s)x + rs = 0.
Since we know that 1 - √2 is a root of the quadratic equation, we can let:
r = 1 + √2
and
s = 1 - √2
Thus, r + s = (1 + √2) + (1 - √2) = 2 and rs = (1 + √2)(1 - √2) = 1 – 2 = -1.
Therefore, the quadratic equation must be x^2 – 2x – 1 = 0.
Answer: D
Step-by-step explanation:
it isn't the first 2 options
I'd say it's y=1/2x - 2
Answer:
The answer is 14 plus y
Step-by-step explanation:
Number A is correct
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