Answer: The blue whale's weight is 150 times heavier than the narwhal's weight.
Step-by-step explanation:
Given: Weight of Blue whale = 
Weight of Narwhal = 
Number of times blue whale's weight is heavier than the narwhal's weight = 
![=\dfrac{3\times10^5}{2\times10^3}\\\\=1.5\times10^{5-3}\ \ \ [\dfrac{a^m}{a^n}=a^{m-n}]\\\\=1.5\times10^2\\\\=1.5\times100=150](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B3%5Ctimes10%5E5%7D%7B2%5Ctimes10%5E3%7D%5C%5C%5C%5C%3D1.5%5Ctimes10%5E%7B5-3%7D%5C%20%5C%20%5C%20%5B%5Cdfrac%7Ba%5Em%7D%7Ba%5En%7D%3Da%5E%7Bm-n%7D%5D%5C%5C%5C%5C%3D1.5%5Ctimes10%5E2%5C%5C%5C%5C%3D1.5%5Ctimes100%3D150)
Hence, the blue whale's weight is 150 times heavier than the narwhal's weight.
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:
61
Step-by-step explanation:
61 is a whole number so you don't need to round it to one.
The vertex of f
(
x
) is at x = −
8/
2
= −
4