Find the arc length for the curve y = 5 x 2 − 1 40 ln x taking P 0 ( 1 , 5 ) as the starting point. If f ( x ) = 5 x 2 − 1 40 ln x , then f ' ( x ) = 1 + [ f ' ( x ) ] 2 = 1 + ( ) 2 = 1 + 100 x 2 − 0.5 + = 100 x 2 + 0.5 + ( ) 2 √ 1 + [ f ' ( x ) ] 2 = . Thus the arc length is given by s ( x ) = ∫ x 1 √ 1 + [ f ' ( t ) ] 2 d t = ∫ x 1 ( ) d t = [ ] x 1 = . For instance, the arc length along the curve from ( 1 , 5 ) to ( 5 , ) is s ( 5 ) = + 1 40 ln 5 − 5 = + ln 5 40 ≈ (rounded to four decimal places).

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

i would say b) is your answer

sorry if im wrong

step-by-step explanation:

easy its decreasing

step-by-step explanation:

because its top point is at the to of the graph on the left side (atleast from the veiw i see it) now if that top point wast in the lower part of the graph it would be increasing hope this have a nice day

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

step-by-step explanation: