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3 years ago

Whoever answers correctly gets 100 points and brainlest

Mathematics

2 answers:

Hatshy [7]3 years ago

7 0

For 1st row

3/x=5/60
3/x=1/12
x=36

For 3 Rd row

x/72=1/12
x=6

For last row

10/x=1/12
x=10(12)
x=120

aleksklad [387]3 years ago

3 0

Answer:

The only complete ordered pair in the given table is:

5 months → $60 saved

Therefore, to determine the relationship between the variables, divide the amount saved by the number of months:

$\implies \sf \dfrac{Amount\:saved}{Number\:of\:months}=\dfrac{\$60}{5}=\$12$

So we can assume that $12 is saved each month.

To complete the table:

Multiply the given number of months by $12 to determine the amount saved.
Divide the given amount saved by $12 to determine the number of months.

⇒ 3 × $12 = $36

⇒ $72 ÷ $12 = 6

⇒ 10 × $12 = $120

Completed table:

$\large\begin{array}{| c | c |}\cline{1-2} \sf Times\:(month) & \sf Amount\:Saved\: (\$) \\\cline{1-2} 3 & 36 \\\cline{1-2} 5 & 60 \\\cline{1-2} 6 & 72 \\\cline{1-2} 10 & 120 \\\cline{1-2} \end{array}$

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Um. It might be 100. I do not now. Sorry

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2 years ago

Diane loves coasters that dip into tunnels during the ride.Her favorite coaster is modeled by h(t)=2t +23t-59t+24. Using rationa

Fiesta28 [93]

Answer:

The possible rational zeros for the function are

±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ±1/2, ±3/2

Step-by-step explanation:

I believe that there is an error in the function with the exponents, it must be:

$h(t) = 2t^{3} + 23t^{2}+59t+24$

If this is the function that you need, then we must use the rational zero theorem. It says that if a polynomial function, written in descending order of the exponents, has integer coefficients, then any rational zero must be of the form ± p/ q, where p is a factor of the constant term and q is a factor of the leading coefficient.

Thus

In this case the constant term is 24 and then

p = ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24

The factor of the leading coefficient is 2, thus

q = ±1, ±2

The possible rational zeros for the function are

±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ±1/2, ±3/2

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3 years ago

Using power series, solve the LDE: (2x^2 + 1) y" + 2xy' - 4x² y = 0 --- - -- -

sattari [20]

We're looking for a solution of the form

$y=\displaystyle\sum_{n\ge0}a_nx^n$

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting these into the ODE gives

$\displaystyle\sum_{n\ge0}\left(\bigg(2(n+2)(n+1)a_{n+2}-4a_n\bigg)x^{n+2}+2(n+1)a_{n+1}x^{n+1}+(n+2)(n+1)a_{n+2}x^n\right)=0$

Shifting indices to get each term in the summand to start at the same power of $x$ and pulling the first few terms of the resulting shifted series as needed gives

$2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0$

Then the coefficients in the series solution are given according to the recurrence

$\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}$

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of $y(0)=y'(0)=1$ , a degree-8 approximation of

$y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}$

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).