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.
Answer:
Step-by-step explanation:
The given initial value problem is;
Let
Differentiating both sides of equation (1) with respect to 
, we obtain:
Differentiating both sides of equation (2) with respect to gives:
From equation (1),
Putting t=0 into equation (2) yields
Also putting t=0 into equation (3)
The system of first order equations is:
Rewrite 15 1/4% as 15.25%. To obtain the desired fraction, divide this 15.25% by 100%:
0.1525
This could be reduced to 61/400, which was obtained by first writing 0.1525 as 1525/10000.
17/30, The peanuts by a dollar per pound (peanuts cost $3, walnuts cost $2), $21, 40072/100 (and I'm not sure how you would reduce it) Good luck!
Answer:
Substitute the given value into the function and evaluate.
Substitute the given value into the function and evaluate.
3 tan (
8
) < − 1 = 0.4216225 < − 1
0 = 0.4216225 < − 1
5 = 0.4216225 < − 1
521 = 0.4216225 < − 1
Step-by-step explanation:
