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).
106.25
the answer would be 106.25 because i was taught to divide 465 by 4 and the subtract 10
Answer:
15a^5b^15
Step-by-step explanation:
you multiply 5 and 3 to get 15, and then when you are multiplying exponents of variables, you actually add them, so a^2 times a^3 is actually a^5, and b^7 times b^8 is b^15. Because these are all multiplied together, you get the answer of 15a^5b^15.
A = area of triangle + area of rectangular + half of the circle area
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triangle area = 1/2 × base × height
triangle area = 1/2 × 3 × 5 = 15/2 = 7.50
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rectangular area = length × width
rectangular area = 6 × 5 = 30
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circle half area = 1/2 × pi × radius^2
circle half area = 1/2 × 3.14 × 3^2
circle half area = 1.57 × 9
circle half area = 14.13
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A = 7.50 + 30 + 14.13
A = 51.63 m^2
(x-h)^2+(y-k)^2=r^2
(h,k) is center
r=radius
(x-2)^2+(y-(-3))^2=2^2
center is (2,-3)
radius is 2
A. (2,-5) is on the circle, not inside
B. (2,-4) is inside
C. (0,-3) is on the circle, not inside
D. (2,0) is on the circle, not inside
answer is B
