Matthew hit 25 home runs last year. This year his home run production dropped 20%. How many home runs did he hit this year? *

Simplify the expression by combining like terms. 22+5−4+7−8

nikklg [1K]

Answer:

22

Step-by-step explanation:

27-4+7-8=

23+7-8=

30-8=

22

3 0

3 years ago

Last week Jonas spent 9 1/6 hours working on his homework over a period of 4 2/3 days. Approximately how many hours a day, on av

Elina [12.6K]

Answer:

55/28 hours/day (approx. 1.96 hours/day)

Step-by-step explanation:

9 1/6 hours / 4 2/3 days

= 55/6 hours / 14/3 days

= 55/6 * 3/14 hours/day

= 55/28 hours/day

(approx. 1.96 hours/day)

3 0

3 years ago

Find the minimum, first quartile, median, third quartile, and maximum of each data set, 12, 10, 11, 7, 9, 10, 5

Sliva [168]

To find the median:
Least to greatest order:

5, 7, 9, 10, 10, 11, 12

10 is the number in the middle. Therefore the median is 10

7 0

3 years ago

Read 2 more answers

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

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

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

3 years ago

Use braces and digits to designate the set of natural numbers.

alina1380 [7]

Answer:

$N = \{1,2,3,4,5,6,.....\}$

Step-by-step explanation:

Required

Determine the sets of natural number

Represent this set with letter N

From definition of natural numbers, they are integer numbers starting from 1;

In other words: 1,2,3,4,5, and so on.

To represent this as a set of N, we have

$N = \{1,2,3,4,5,6,.....\}$

<em>The ..... means that the counting still continues</em>

6 0

3 years ago