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2 years ago

10

Sam confirmed there will be 192 guests at his party. He thinks each guest will eat

1 answer:

vitfil [10]2 years ago

7 0

Answer:

Sam used estimation. He rounded the amount of guests (192) to the nearest hundred (200).

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What does the range and mean tell you about data?

Nady [450]

The approximate middle value

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3 years ago

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Question 6 (10 points)

larisa [96]

Answer:

22

Step-by-step explanation:

The ratio tells us 2/5 of the swimmers are indoor swimmers.

55(2/5)=22

6 0

1 year ago

Ming's team cleans 3 animals per hour after an oil

erik [133]

Answer: 2 hours

Step-by-step explanation:

$Mingn's \ team \\\\ 3\ animals -1 h \\\\ 12 \ animals -? h \\\\ ?=\dfrac{12}{3} =\boxed{4 \ hours}$ $Nina's \ team \\\\ 6\ animals -1 h \\\\ 12 \ animals -? h \\\\ ?=\dfrac{12}{6} =\boxed{2 \ hours}$

$\rm Difference: \\\\4-2=\bf 2 h$

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3 years ago

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We would like to use the power series method to find the general solution to the differential equation d 2y dx2 − 4x dy dx + 12y

Feliz [49]

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

$\dfrac{\mathrm dy}{\mathrm dx}=\displaystyle\sum_{n\ge1}na_nx^{n-1}\implies4x\dfrac{\mathrm dy}{\mathrm dx}=4\sum_{n\ge1}na_nx^n=4\sum_{n\ge0}na_nx^n$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting into the ODE

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-4x\dfrac{\mathrm dy}{\mathrm dx}+12y=0$

gives

$\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}-4na_n+12a_n\bigg)x^n=0$

so that the coefficients of the series are given according to

$\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\dfrac{4(n-3)a_n}{(n+2)(n+1)}&\text{for }n\ge0\end{cases}$

We can shift the index in the recursive part of this definition to get

$a_n=\dfrac{4(n-5)a_{n-2}}{n(n-1)}$

for $n\ge2$ . There's dependency between coefficients that are 2 indices apart, so we can consider 2 cases:

If $n=2k$ , where $k\ge0$ is an integer, then

$k=0\implies n=0\implies a_0=a_0$

but since $y(0)=0$ , we have $a_0=0$ and $a_{2k}=0$ for all $k\ge0$ .

If $n=2k+1$ , then

$k=0\implies n=1\implies a_1=a_1$

$k=1\implies n=3\implies a_3=\dfrac{4(-2)a_1}{3\cdot2}=-\dfrac43a_1$

$k=2\implies n=5\implies a_5=0$

and so $a_{2k+1}=0$ for all $k\ge2$ . If $y'(0)=1$ , we then have $a_1=1$ and $a_3=-\dfrac43$ .

So the ODE has solution

$y(x)=\displaystyle\sum_{k\ge0}(a_{2k}x^{2k}+a_{2k+1}x^{2k+1})\implies\boxed{y(x)=x-\dfrac43x^3}$

8 0

3 years ago

3(x-1) - 8 = 4(1+x) +5

ryzh [129]

For this case we must simplify the following expression and find the value of the variable "x":

$3 (x-1) -8 = 4 (1 + x) +5$

We apply distributive property to the terms of parentheses:

$3x-3-8 = 4 + 4x + 5$

We add similar terms:

$3x-11 = 9 + 4x$

We subtract 4x on both sides:

$3x-4x-11 = 9\\-x-11 = 9$

We add 11 to both sides:

$-x = 11 + 9\\-x = 20\\x = -20$

Answer:

$x = -20$

8 0

4 years ago