Answer:
2. a and b only.
Step-by-step explanation:
We can check all of the given conditions to see which is true and which false.
a. f(c)=0 for some c in (-2,2).
According to the intermediate value theorem this must be true, since the extreme values of the function are f(-2)=1 and f(2)=-1, so according to the theorem, there must be one x-value for which f(x)=0 (middle value between the extreme values) if the function is continuous.
b. the graph of f(-x)+x crosses the x-axis on (-2,2)
Let's test this condition, we will substitute x for the given values on the interval so we get:
f(-(-2))+(-2)
f(2)-2
-1-1=-3 lower limit
f(-2)+2
1+2=3 higher limit
according to these results, the graph must cross the x-axis at some point so the graph can move from f(x)=-3 to f(x)=3, so this must be true.
c. f(c)<1 for all c in (-2,2)
even though this might be true for some x-values of of the interval, there are some other points where this might not be the case. You can find one of those situations when finding f(-2)=1, which is a positive value of f(c), so this must be false.
The final answer is then 2. a and b only.
-z³ + 5k^6 + z³ -10k^6
(-z³ cancels out with z³)
5k^6 -10k^6
(then subtract)
Answer is -5k^6
Answer:
x - 3
Step-by-step explanation:
We know that the area of a square is length times width.
We also know that a quadratic can be factored:
(x - 3)²
So if each side is (x - 3), then when we do length times width, we get x² - 6x + 9. So our answer is choice D.
Simply add the two dosages together, (0.15+0.025) and the answer is 0.175 :)
Answer:
Center: (3,5)
Radius: 2
Step-by-step explanation:
x² + y² - 6x - 10y + 30 = 0
Put equation into center-radius form. First regroup terms:
(x²-6x) + (y²-10y) = -30
Complete the squares:
(x²-6x+3²) + (y²-10y+5²) = -30 + 3² + 5²
(x-3)² + (y-5)² = 4 = 2²
Center: (3,5)
Radius: 2
