The local minimum of function is an argument x for which the first derivative of function g(x) is equal to zero, so:
g'(x)=0
g'(x)=(x^4-5x^2+4)'=4x^3-10x=0
x(4x^2-10)=0
x=0 or 4x^2-10=0
4x^2-10=0 /4
x^2-10/4=0
x^2-5/2=0
[x-sqrt(5/2)][x+sqrt(5/2)]=0
Now we have to check wchich argument gives the minimum value from x=0, x=sqrt(5/2) and x=-sqrt(5/2).
g(0)=4
g(sqrt(5/2))=25/4-5*5/2+4=4-25/4=-9/4
g(-sqrt(5/2))=-9/4
The answer is sqrt(5/2) and -sqrt(5/2).
La fille a mudane guah garcon la homme la femme. 6/9
5/1, 15/3, and 10/2 are all valid answers to that question.
Let's create this right triangle. We will call it triangle ABC with angle B as the right angle, angle A being the larger of the angles A and C. If the ratio is 5:4, then it is A:C. 5+4 = 9. The measure of the right angle, angle ABC = 90, and 90 divided by 9 is 10. So angle A is 5 * 10 which is 50 degrees, and angle C is 4 * 10 which is 40 degrees. The smallest angle, angle C, is 40 degrees.
Just like you drew those dotted lines. You know how to find the areas of rectangles which is A=length x width
The answer would be 20cm^2+6cm^2+18cm^2=44 cm^2
