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3 years ago

What is negative six one third equal negative nine three fourths

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

8 0

Answer:

Do you have a picture of the problem so I can see it

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Find the length of UC on the attached diagram. Thank you!

ValentinkaMS [17]

<h2>Answer </h2>

The length of UC is 18

<h2>Explanation </h2>

First we are going to find the length of JN; then we are subtracting from it the length of JU plus the length of CN.

We can infer from our picture that JN is 82 + 105, so JN = 187

We can also infer that JU = JH + HU

JU = 22 + 96

JU = 118

We can also infer that CN = 51

Now we can fin the length of UC:

$UC=JN-(JU+CN)$

$UC=187-(118+51)$

$UC=18$

We can conclude that the length of UC is 18.

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3 years ago

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1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

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

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

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

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

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3 years ago

Write an equation of the line that passes through the given points. (-1,7) (3,-9)

Novay_Z [31]

Answer:

x=13

hope you do well

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3 years ago

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Need heals please ❤️

Neko [114]

B: Even

B is the answer

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3 years ago

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Arthemeatic sequence

inn [45]

Answer:

Step-by-step explanation:

its going down by 11

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2 years ago