Answer:
American Crows can be considered partially migratory. That is, some populations migrate, others are resident, and in others only some of the crows migrate. Crows in the southern parts of their range appear to be resident and not migrate. They may make some changes in their use of space at this time, spending more time off the territory to forage and roost. Crows migrate out of the northern most parts of their range. It has been stated that crows migrate out of those areas where the minimum January temperature averages 0 ° F. Certainly crows leave the northern Great Plains in the fall, leaving Saskatchewan and Alberta to winter in the lower Plains states of Nebraska, Kansas, and Oklahoma (Kalmbach, E. R., and S. E. Aldous. 1940. Winter banding of Oklahoma crows. Wilson Bull. 52: 198-206). Crows can be seen crossing the Great Lakes in spring and fall, and these birds undoubtedly are migrating to and from parts of Canada.
Step-by-step explanation:
because They may make some changes in their use of space at this time, spending more time off the territory to forage and roost. Crows migrate out of the northern most parts of their range. It has been stated that crows migrate out of those areas where the minimum January temperature averages 0 ° F.
Answer: 220 feet tall
Step-by-step explanation:
1. Draw!
2. Use trigonometric ratios to solve for the height (Remember SOH CAH TOA)
One A
y = e^x
dy/dx = e^x The f(x) = the differentiated function. Any value that e^x can have, the derivative has the same value. x is contained in all the reals.
One B
y = x*e^x
y' = e^x + xe^x Using the multiplication rule.
You want the slope and the value of the of y to be the same. The slope is y' of the tangent line
xe^x = e^x + xe^x
e^x = 0
This happens only when x is very "small" like x = - 4444444
y = e^x * ln(x) Using the multiplication rule again, we need the slope of the line with is y'
y1 = e^x
y1' = e^x
y2 = ln(x)
y2' = 1/x
y' = e^x*ln(x) + e^x/x So at x = 1 the slope of the line =
y' = e^1*ln(1) + e^1/1
y' = e*0+e = e
y = mx + b
y = ex + b
to find b we use y= e^x ln(x)
e^x ln(x) = e*x + b
e^1 ln(1) = e*1 + b
ln(1) = 0
0 = e + b
b = - e
line equation and answer.
y = e*x - e
Find the critical points of 
:
All three points lie within 
, and 
takes on values of
Now check for extrema on the boundary of . Convert to polar coordinates:
Find the critical points of 
:
where 
is any integer. There are some redundant critical points, so we'll just consider 
, which gives
which gives values of
So altogether, has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).
