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2 years ago

A line passes through the point (4, -6) and has a slope of -3/4. Which is the

Mathematics

1 answer:

soldi70 [24.7K]2 years ago

7 0

Answer:

I have not dealt with slope for a while but if its in the form y=mx+b then the x is the slope so it would be -3/4x and b is the y intercept. the y in the cords is -6

Step-by-step explanation:

y= -3/4x -6

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60 students want lemonade and 16 students want iced tea. What is the ratio of the number of students who want iced tea to the nu

Burka [1]

Answer:

16:60 = 4/15

Step-by-step explanation:

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2 years ago

Find the measure of VU Can someone help me?? I need to show the work

borishaifa [10]

Answer:

Step-by-step explanation:

lets solve the problem 7x+2=3x+10

subtract 3x from 7x

4x+2=10

subtract 2 from 10

4x=8

divide both sides by 4

x=2

to find VU plug in 2 for x so...

3(2)+10

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I hope this helped you :)

7 0

2 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:

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

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2 years ago

Find the value of x plsss the people that answerd this got it wrong.

liubo4ka [24]

X=145. If its wrong lmk

6 0

2 years ago

Read 2 more answers

Which expressions are equivalent to Negative 9 (two-thirds x + 1)? Check all that apply.

Sav [38]

Answer:

$-9(\frac{2}{3}x) + 9(1)\\ \\-6x + 9$