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2 years ago

Help please?! This is graphing rational functions

Mathematics

1 answer:

guapka [62]2 years ago

7 0

Answer:

D: (-∞,2) (2,∞)

R:(-∞,0),(0,∞)

Step-by-step explanation:

From the graph we can see that the graph is approaching 0 horizontally and 2 vertically but not crossing so we can assumer these are the asymptotes. So the domain goes all the way from -∞ to 2 but does not cross 2 so we have to stop there and create another bracket for 2 to ∞. It is the same process for the range but in the y direction.

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Please explain your answer. THX!

Elden [556K]

Answer:

see explanation

Step-by-step explanation:

(3)

Given cosΘ = - $\frac{4}{5}$

Then by Pythagoras' theorem the third side is 3 ( 3,4, 5 triangle )

Since Θ in second quadrant then sinΘ > 0

sinΘ = $\frac{3}{5}$

Using the trigonometric identity

sin2Θ = 2sinΘcosΘ, then

sin2Θ = 2 × $\frac{3}{5}$ × - $\frac{4}{5}$ = - $\frac{24}{25}$

(4)

Using the trigonometric identity

cos(x - y) = cosxcosy + sinxsiny

note cos15° = cos(45 - 30)°

cos(45 - 30) = cos45cos30 + sin45sin30

= ( $\frac{\sqrt{2} }{2}$ × $\frac{\sqrt{3} }{2}$ ) + ( $\frac{\sqrt{2} }{2}$ × $\frac{1}{2}$ )

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3 years ago

Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

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