Calculation of relative maxima and minima of a function f (x) in a range [a, b]:
We find the first derivative and calculate its roots.
We make the second derivative, and calculate the sign taken in it by the roots of the first derivative, and if:
f '' (a) <0 is a relative maximum
f '' (a)> 0 is a relative minimum
Identify intervals on which the function is increasing, decreasing, or constant. G (x) = 1- (x-7) ^ 2
First derivative
G '(x) = - 2 (x-7)
-2 (x-7) = 0
x = 7
Second derivative
G '' (x) = - 2
G '' (7) = - 2 <0 is a relative maximum
answer:
the function is increasing at (-inf, 7)
the function is decreasing at [7, inf)
Order from least to greatest
-9/2, - 3, 1/2, sqrt(6), 4.2
answer
D.-9/2, - 3, 1/2, sqrt(6), 4.2
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sqrt(10) is irrational number and real numbers
answer is B.irrational number and real numbers
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square root (81) = 9
square root (100) = 10
square root (93) is between 9 and 10
so answer
9< sqrt(93) < 10
Since AED forms a straight line, all involved angles sum to 180°. Therefore AEB + BED = 180. Also, since EC bisects BED, BEC = CED, and BED = 2× CED. Now to substitute the first equation:
AEB + BED = 180
AEB + 2×CED = 180
11x-12 + 2(4x+1) = 180
11x-12+8x+2 = 180
19x-10 = 180
19x-10+10 = 180+10
19x = 190
x = 10
So what is m<AEC?? It is the sum of AEB + BEC, and since BEC = CED we can say that:
AEC = AEB + CED
AEC = 11x-12 + 4x+1 = 15x-11 = 15(10)-11 = 150-11
m<AEC = 139°
Answer:
Rise is a vertical line from 0 to 4
Run is a horizontal line from 0 to 4
Slope of a line is called the gradient.
Additional information:
Gradient =rise/run
= 4/4
= 1
Y=mx +c
Y=1x +4
Y= x+4
But the slope show a negative correlation, so gradient will be negative & it's y = - x + 4.
BTW the rise and run could also be from 4 on the y-axis and down to the 4 on the x-axis.
So, there is two possible triangles you could make.
As a result, the rise is a straight line going up or down ( vertical ) and the run is a straight line going across from left to right (horizontal).
