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3 years ago

1) A Rectangle has a length of 4x+2 and a width of 7x+6. The

Mathematics

1 answer:

Dima020 [189]3 years ago

7 0

B should be the correct answer

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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

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

7 0

3 years ago

PLS HALP ASAP =w=!!!! (will mark u teh brainliest if you explain the problem properly and clearly, 20 pts uwu!)

Oksanka [162]

N = 3 i = 2
2 x 2 = 4
3 x 3 = 9
4 + 9 = 13
13 = area

7 0

3 years ago

A news reporter says that 28% of highschool students pack their lunch. your highschool has 600 students.

Vika [28.1K]

Answer:

168

Step-by-step explanation:

0.28 * 600

3 0

3 years ago

Read 2 more answers

Harper plans to purchase her school pictures. The prints cost $108 plus any add-ons she wants to purchase. If she saves $18 per

Lemur [1.5K]

Answer:

It will take 6 weeks

Step-by-step explanation:

Weeks = x

$18x = $108

X = $108/$18

X = 6

So 6 weeks

4 0

3 years ago

Match the expression to the exponent property that you use first to simplify the expression.

IRINA_888 [86]

Step-by-step explanation:

$\dfrac{a^m}{a^n}=a^{m-n}\to\dfrac{h^\frac{3}{2}}{h^\frac{4}{3}}=h^{\frac{3}{2}-\frac{4}{3}}=h^{\frac{(3)(3)}{(2)(3)}-\frac{(2)(4)}{(2)(3)}}=h^{\frac{9}{6}-\frac{8}{6}}=h^{\frac{1}{6}}\\\\(a^m)^n=a^{mn}\to\bigg(p^\frac{1}{4}\bigg)^\frac{2}{3}=p^{\left(\frac{1}{4}\right)\left(\frac{2}{3}\right)}=p^\frac{2}{12}=p^\frac{1}{6}\\\\a^m\cdot a^n=a^{m+n}\to z^\frac{3}{4}\times z^\frac{5}{6}=z^{\frac{3}{4}+\frac{5}{6}}=z^{\frac{(3)(3)}{(4)(3)}+\frac{(5)(2)}{(6)(2)}}=z^{\frac{9}{12}+\frac{10}{12}}=z^\frac{19}{12}$

$\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\to\bigg(\dfrac{x^2}{y}\bigg)^\frac{1}{3}=\dfrac{\left(x^2\right)^\frac{1}{3}}{y^\frac{1}{3}}=\dfrac{x^{(2)\left(\frac{1}{3}\right)}}{y^\frac{1}{3}}=\dfrac{x^\frac{2}{3}}{y^\frac{1}{3}}$

3 0

3 years ago