The coordinate grid shows the cost of different numbers of book marks. How many cents do 4 bookmarks cost?

Mathematics

1 answer:

Mandarinka [93]2 years ago

5 0

Answer:

70 cents

Step-by-step explanation:

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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

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

4 0

1 year ago

insens350 [35]

To start, the equation has to be converted to slope-intercept form, or $y=mx+b$ . Start by subtracting x from both sides; $x-x+2y=4$ , $2y=4-x$ . Switch 4 and -x around to conform to the formula; $2y=-x+4$ . Then divide by 2 to isolate y; $y=-\frac{1}{2}x+2$ . This is the final equation. Because the slope of a line perpendicular is always the opposite reciprocal (i.e. The opposite reciprocal of 4 is $-\frac{1}{4}$ , the opposite reciprocal of $-\frac{1}{5}$ is 5) the opposite reciprocal of $-\frac{1}{2}$ is 2. So the equation of the line is $y=2x$ , and the answer is A.