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3 years ago

Pls help me it's due today

Mathematics

1 answer:

Dimas [21]3 years ago

8 0

Answer:

r³ + 15r² - 10r - 4

Step-by-step explanation:

First, note you can remove the brackets, in this case they contribute nothing at all.

Then, group the like terms and add them, e.g., we have 3r³ and -2r², they add up to r³.

Finally, order them from greatest power of r to smallest power of r.

That's all!

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Of the students at I.S. 59, 230 signed up for the school picnic. That was 50% of all students in the school. How many students a

Nastasia [14]

Answer: 460 students

Step-by-step explanation:

Based on the information given in the question, let the total number of students that attend I.S. 59 be represented by x.

Since we've been given the information that 230 signed up for the school picnic which was 50% of all students in the school. Then, the students in the school will be:

= 50% of x = 230

0.5 × x = 230

0.5x = 230

x = 230/0.5

x = 460

Therefore, 460 students attend the school.

4 0

3 years ago

A set of data was published in the fall 2019 Phi Kappa Phi Forum regarding the performance of Major League Baseball (MLB) umpire

const2013 [10]

Answer:

The answer is "1.30".

Step-by-step explanation:

Proportion of population $=0.20$

Ratio sample $=\frac{70}{300}=\frac{7}{30}=0.23$

Statistics of the test:

$\to Z=\frac{(0.23-0.2)}{\sqrt{(\frac{0.2\times(1-0.2)}{300})}}\\\\$

$=1.30$

6 0

3 years ago

X=4y-5<br> 2x+3y=23<br> Solve using the substitution method

jenyasd209 [6]

Answer:

x =7, y = 3

Step-by-step explanation:

We substitute the value of x in terms of y (given) into the equation.

Then we solve for y and plug in the value of y back into the equation for the value of x to find x.

3 0

3 years ago

Explain this strategy: 1.4 to the power of 4 times .75 squared equals 2.1609

yaroslaw [1]

What do u think the answer is

3 0

3 years ago

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

Pls help me it's due today​

Pls help me it's due today