Find a curve y=f(x) with the following properties (i) y"=6x and (ii) its graph passes through the point (1,1) and has a horizontal tangent there.

First solve the function by integrating the 2nd derivative of the function two times. Since the y’’ = 6x

Integrating

Y’ =∫ 6x

y’ = 6 (x^2)/2

y’ = 3(x^2) + c

integrating it further to solve the function

y =∫ 3(x^2) + c

y = 3(x^3)/3 + c1

y = (x^3) + c1

to solve c1

substitute the point (1,1)

1 = (1^3) + c1

C1 = 0

So the function is

y = x^3

answer:  

step-by-step explanation:

odd numbers are only divisible by 1 and itself.

even numbers are divisible by more than 1 and itself.

for example 6 is even because it is divisible by more than 1 and its self. 1 x 6= 6 ; 2 x 3= 6

and 7 is odd because only 7 x 1 equals seveb

it's 100

step-by-step explanation:

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

  n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the  14th  row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100

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