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stiks02 [169]
3 years ago
5

You can buy 5 apples for $6.25 or 8 apples for $10.24. Which is the better deal? Explain.

Mathematics
2 answers:
balu736 [363]3 years ago
5 0

Answer:

8 apples for $10.24 as your getting more apples for almost the same price.

Step-by-step explanation:

$6.25 divided by 5 is $1.25 so that would be $1.25 per apple

$10.24 divided by 8 is $1.28 so that would be $1.28 per apple

kakasveta [241]3 years ago
5 0

Answer:

5 apples for 6.25 dollars is the better deal.

Step-by-step explanation:

I set both deals as a ratio and cross multiplied them with X/1 to get how much each apple would cost per one apple. The first deal ended up being cheaper with 1.25 dollars per apple. The more expensive deal was 1.28 dollars per apple. Therefore 5 apples for 6.25 dollars is the better deal.

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Xy''+2y'-xy by frobenius method
aalyn [17]
First note that x=0 is a regular singular point; in particular x=0 is a pole of order 1 for \dfrac2x.

We seek a solution of the form

y=\displaystyle\sum_{n\ge0}a_nx^{n+r}

where r is to be determined. Differentiating, we have

y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}
y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}

and substituting into the ODE gives

\displaystyle x\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-x\sum_{n\ge0}a_nx^{n+r}=0
\displaystyle \sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle \sum_{n\ge0}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge2}a_{n-2}x^{n+r-1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}\bigg((n+r)(n+r+1)a_n-a_{n-2}\bigg)x^{n+r-1}=0

The indicial polynomial, r(r+1), has roots at r=0 and r=-1. Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.

When r=0, we have the recurrence

a_n=\dfrac{a_{n-2}}{(n+1)(n)}

valid for n\ge2. When n=2k, with k\in\{0,1,2,3,\ldots\}, we find

a_0=a_0
a_2=\dfrac{a_0}{3\cdot2}=\dfrac{a_0}{3!}
a_4=\dfrac{a_2}{5\cdot4}=\dfrac{a_0}{5!}
a_6=\dfrac{a_4}{7\cdot6}=\dfrac{a_0}{7!}

and so on, with a general pattern of

a_{n=2k}=\dfrac{a_0}{(2k+1)!}

Similarly, when n=2k+1 for k\in\{0,1,2,3,\ldots\}, we find

a_1=a_1
a_3=\dfrac{a_1}{4\cdot3}=\dfrac{2a_1}{4!}
a_5=\dfrac{a_3}{6\cdot5}=\dfrac{2a_1}{6!}
a_7=\dfrac{a_5}{8\cdot7}=\dfrac{2a_1}{8!}

and so on, with the general pattern

a_{n=2k+1}=\dfrac{2a_1}{(2k+2)!}

So the first indicial root admits the solution

y=\displaystyle a_0\sum_{k\ge0}\frac{x^{2k}}{(2k+1)!}+a_1\sum_{k\ge0}\frac{x^{2k+1}}{(2k+2)!}
y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}
y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}

which you can recognize as the power series for \dfrac{\sinh x}x and \dfrac{\cosh x}x.

To be more precise, the second series actually converges to \dfrac{\cosh x-1}x, which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When r=-1, we may seek a second solution of the form

y=cy_1\ln x+x^{-1}\displaystyle\sum_{n\ge0}b_nx^n

where y_1=\dfrac{\sinh x+\cosh x-1}x. Substituting this into the ODE, you'll find that c=0, and so we're left with

y=x^{-1}\displaystyle\sum_{n\ge0}b_nx^n
y=\dfrac{b_0}x+b_1+b_2x+b_3x^2+\cdots

Expanding y_1, you'll see that all the terms x^n with n\ge0 in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form y_2=\dfrac1x. Adding this to y_1, we end up with just \dfrac{\sinh x+\cosh x}x.

This means the general solution for the ODE is

y=C_1\dfrac{\sinh x}x+C_2\dfrac{\cosh x}x
3 0
3 years ago
What is 6.9 to the nearest 10th
Dvinal [7]
The number 6.9 is already rounded to the nearest tenth.
5 0
3 years ago
0.2x-1=-1 what is x in the equation
Alenkinab [10]
I believe the answer is x=0
8 0
3 years ago
I need help with this (Image Attached)
NemiM [27]

Answer:

55,98,27

Step-by-step explanation:

To solve for the angles we must first find y with this equation

3y+5+2y-7=153

We can simplify this

5y-2=153

153+2=155

155/5=31

y=31

We now know y and can plug it into the angles

<A=2y-7

2(31)-7

62-7

55

<A=55

<B=3y+5

3(31)+5

93+5

98

<B=98

Now time for the last angle

Since angle BCA and DCB are next to each other they add up to 180

153+x=180

180-153=27

x=27

<BCA=27

4 0
3 years ago
Nikita has to get eye surgery that will cost $915. She can’t afford that right now so the eye center says she can pay $75 now an
Vesnalui [34]

Answer:

She will have paid $168 not including the original $75 and $243 with.

6 0
3 years ago
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