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:
Area of the land A - Length * Width
Given
A = 1 3/4 mi²
Length L = 2 1/3 miles
Required
Width of the property
Substitute into the formula;
A = LW
W = A/L
W = 1 3/4/(2 1/3)
W = 7/4 ÷ 7/3
W = 7/4 × 3/7
W = 3/4 miles
Hence the width of the property is 3/4 miles
For 6 points, answering 5 questions is a rip-off.
However, I shall answer 7.
haha sike get noob
Answer:
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Step-by-step explanation:
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