3 years ago

Find the length of the curve y=3x^3/2 from x=0 to x=5/9

1 answer:

4 0

$y=3x^{3/2}\implies y'=\dfrac92x^{1/2}$

The length of the curve is

$L=\displaystyle\int_0^{5/9}\sqrt{1+\left(\dfrac92x^{1/2}\right)^2}\,\mathrm dx=\int_0^{5/9}\sqrt{1+\dfrac{81}4x}\,\mathrm dx=\dfrac{335}{243}$

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The coordinates of the endpoints of directed line sergeant ABC are A(-8,7) and C(7,-13). If AB:BC = 3:2 then the coordinates of

cestrela7 [59]

The correct answer is A) (1,-5)

Further explanation:

Given points are:

A(-8,7)=(x1,y1)

C(7,13)=(x2,y2)

IT is also given that

AB:BC=3:2

Which means that B divides the line segment in 3:2

Here,

m=3

n=2

To find the coordinates of B

$x_B=\frac{mx_2+nx_1}{m+n}\\ = \frac{(3)(7)+(2)(-8)}{3+2}\\=\frac{21-16}{5}\\=\frac{5}{5}\\=1\\y_B=\frac{my_2+ny_1}{m+n}\\=\frac{(3)(-13)+(2)(7)}{3+2}\\=\frac{-39+14}{5}\\=\frac{-25}{5}\\=-5$