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.
The answer for this question I don’t think could be answered by anyone but yourself. To help you figure this out, just simply think about all of the math courses you’ve had previous to your current course and think about which course you did best in.
Answer: 25
Step-by-step explanation:
Subtract 46.50 - 46.25 and you’ll get 25
Answer:
-0.8x + 4.8y + 16
Step-by-step explanation:
4(0.5x+2.5y-0.7x-1.3y+4) = 4( 0.5x - 0.7x + 2.5y - 1.3y + 4)
= 4( -0.2x + 1.2y + 4)
= 4*(-0.2x) + 4 *1.2y + 4*4
= -0.8x + 4.8y + 16
5.4 because the 9 bumps it to 5.35.
And the 5 bumps it to 5.4.
