<span>If the half-life of the drug is 2 hours, then in 2 hours 0.6 mg of the drug becomes 0.3 mg. In two more hours (4 hours total) it is 0.15 mg, 2 more hours (6 hours) .075 mg. In two more hours (8 hours) it is .0375 mg, rounded to 0.04. The answer is 8 hours.</span>
So whatever happened does not matter, They are only asking the volume of liquid left in A at the end. Therefore :
Calculation :
B + 120 = 4A
B = 4A - 120 ---------- 1
A - 120 = B ------------ 2
Subtract 2 from 1
B - B = 4A - 120 - A - 120
0 = 3A - 240
240 = 3A
240 ÷ 3 = A
8 = A
Answer :
A = 8
If my memory serves me well, the only wtatement which is true is the last one: <span>D) All linear equation and all quadratic equations are polynomial equations.</span>
Answer:
-The answer is
.
Step-by-step explanation:
When you divide numbers/variables with exponents, the exponents will subtract each other:

Starting off with the polynomial in standard form would be extremely difficult, but we can construct one fairly easily with the zeroes we've been given.
We know from the given zeroes that our function has the value 0 when x = 1, x = -2, and x = 2. Manipulating each equation, we can rewrite them as x - 1 = 0, x + 2 = 0, and x - 2 = 0. To construct our polynomial, we simply use all three of the expressions on the left side of the equation as factors and multiply them together, obtaining:

Notice that we can easily obtain each our three zeroes by dividing both sides by the two other factors. From here, we just need to expand the left-hand side of the equation. I'll show the work required here:
=0\\ (x^2-x+2x-2)(x-2)=0\\ (x^2+x-2)(x-2)=0\\ (x^2+x-2)x-(x^2+x-2)2=0\\ x^3+x^2-2x-(2x^2+2x-4)=0\\ x^3+x^2-2x-2x^2-2x+4=0\\ x^3+(x^2-2x^2)+(-2x-2x)+4=0\\ x^3-x^2-4x+4=0\\](https://tex.z-dn.net/?f=%28x-1%29%28x%2B2%29%28x-2%29%3D0%5C%5C%0A%5Cbig%5B%28x-1%29x%2B%28x-1%292%5Cbig%5D%28x-2%29%3D0%5C%5C%0A%28x%5E2-x%2B2x-2%29%28x-2%29%3D0%5C%5C%0A%28x%5E2%2Bx-2%29%28x-2%29%3D0%5C%5C%0A%28x%5E2%2Bx-2%29x-%28x%5E2%2Bx-2%292%3D0%5C%5C%0Ax%5E3%2Bx%5E2-2x-%282x%5E2%2B2x-4%29%3D0%5C%5C%0Ax%5E3%2Bx%5E2-2x-2x%5E2-2x%2B4%3D0%5C%5C%0Ax%5E3%2B%28x%5E2-2x%5E2%29%2B%28-2x-2x%29%2B4%3D0%5C%5C%0Ax%5E3-x%5E2-4x%2B4%3D0%5C%5C)
So, in standard form, our cubic polynomial would be