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3 years ago

11

Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

2 answers:

n200080 [17]3 years ago

8 0

I answered your question before 8.5/2 = 4.25 meaning
17/4= 4.25 which is your answer

Yanka [14]3 years ago

3 0

Answer:

The answer is A. k = 4.25.

Step-by-step explanation:

To find proportionality, solve for 8.5 divided by 2 and 17 divided by 4.

Both of those answers are 4.25, so that would be the final answer.

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>

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I need answers for all! thank you will mark brainlist!! please help me i’m begging you <3

Darina [25.2K]

Step-by-step explanation:

y = 2/5 x - 9/5

when x = -1,

the value of y = 2/5(-1) - 9/5

= -2/5 - 9/5 = -11/5

when x = 0

=> y = 2/5(0) -9/5

= 0-9/5 = -9/5

when x=1

=> y = 2/5(1) - 9/5

= 2/5 - 9/5 = 7/5

6 0

3 years ago

Which of the following shows the prime factorization of 72 using exponential notation?

Luba_88 [7]

Answer:

2^3 .3^2

Step-by-step explanation:

if 2×2×2 = 8

and 3×3=9 then 8×9=72

6 0

3 years ago

Read 2 more answers

Compare 900 cm 9000 mm

dmitriy555 [2]

9000mm is equivalent to 900cm

3 0

3 years ago

Hello i need help with this plz

gayaneshka [121]

Answer: Sorry i only could so #3

32/30

Step-by-step explanation:

7 0

2 years ago

Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a

Crank

I'll assume the ODE is actually

$y''+(x-2)y'+y=0$

Look for a series solution centered at $x=2$ , with

$y=\displaystyle\sum_{n\ge0}c_n(x-2)^n$

$\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n$

$\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n$

with $c_0=y(2)=2$ and $c_1=y'(2)=0$ .

Substituting the series into the ODE gives

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0$

$\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}$

If $n=2k$ for integers $k\ge0$ , then

$k=0\implies n=0\implies c_0=c_0$

$k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}$

$k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}$

$k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}$

and so on, with

$c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}$

If $n=2k+1$ , we have $c_{2k+1}=0$ for all $k\ge0$ because $c_1=0$ causes every odd-indexed coefficient to vanish.

So we have

$y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}$

Recall that

$e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}$

The solution we found can then be written as

$y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k$

$\implies\boxed{y(x)=2e^{-(x-2)^2/2}}$

6 0

3 years ago