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3 years ago

Sketch the curve y=x(x-2)^3 showing any stationary points and points of inflection

Mathematics

2 answers:

ycow [4]3 years ago

6 0

Answer:

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elixir [45]3 years ago

3 0

Answer:

Step-by-step explanation:

hello :

let : y=f(x)

f(x)==x(x-2)^3 so : f'(x) = 1.(x-2)^3+3x(x-2)²

f'(x) = (x-2)²(x-2+3x)

f'(x) = (x-2)²(4x-2)

f'(x) =0 : x=2 or x=1/2

the minumum is (1/2 ,f(2))

the inflection points use f"(x) = 0.......continu

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Two roller coasters are vying for the title of the steepest drop the first ride has a hill that drops 140 feet in a 40 foot run

muminat

Answer:

They have the same slope

Step-by-step explanation:

Given

Coaster 1:

$Drops = 140ft$

$Run = 40ft$

Coaster 2:

$Drops = 105ft$

$Run = 30ft$

First, we calculate the slope (m) of both coasters

$m = \frac{Rise}{Run}$

$m = \frac{Drops}{Run}$

For coaster 1:

$m = \frac{140ft}{40ft}$

$m = \frac{140}{40}$

$m = 3.5$

For coaster 1:

$m = \frac{105ft}{30ft}$

$m = \frac{105}{30}$

$m = 3.5$

$m = m=>3.5 = 3.5$

By comparison, they have the same slope

8 0

3 years ago

GEOMETRY PLEASE HELP ME URGENT

Basile [38]

Converse (switch p and q)
If an angle is obtuse, then it measures 128°

This is false (a 127° angle is obtuse, but it does not measure 128°)

_____________________________________________________________

Inverse (negations of p and q)
If an angle does not measure 128°, then it is not obtuse

This is false (a 127° angle does not measure 128°, but it is obtuse)

_____________________________________________________________

Contrapositive (negations of p and q, then switch their places)

If an angle is not obtuse, it does not measure 128°

This is true (any 128° is obtuse; no exceptions)

5 0

3 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$