Find the arc length of the curve below on the given interval. y equals 2 x Superscript 3 divided by 2 on [0 comma 9 ]

d=54.93

Step-by-step explanation:

From exercise we have y=2x^{3/2} on interval [0,9]

We calculate the arc length of the curve  on the given interval.

We have the formula:

d=\int\limits^a_b {\sqrt{1+f'(x)^2}} \, dx

we get

f(x)=y=2x^{3/2}

f'(x)=y'=3x^{1/2}

d=\int\limits^9_0 {\sqrt{1+(3x^{1/2})²}\, dx

d=\int\limits^9_0 {\sqrt{1+9x}}\, dx

d=[\frac{2·(9x+1)^{3/2}}{27}]_0^9

d=\frac{2·(82^{3/2}-2)}{27}

d=54.93

We use the site geogebra.org to drawn a graph.

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)