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3 years ago

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HELP HELP HELP!!!!! PLEASE ITS DUE TODAY YOU CAN DO EITHER OR BOTH ( I need both but any help is accepted with thanks and gradit

ude) WILL GIVE BRAINLIST TO WHOEVER GETS BOTH
(PLEASE PLEASE PLEASE ANY HELP!!!!)

1 answer:

Free_Kalibri [48]3 years ago

3 0

Answer: what do you need help with?

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try typing a question pls

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7 less than the quotient of a number 5 and w in a algebraic expression.

Aleksandr-060686 [28]

Answer:

5/w -7

Step-by-step explanation:

quotient means division

5/w

less than means it comes after

5/w -7

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3 years ago

Read 2 more answers

Is 4x^2 + 20x^2 + 25 a perfect square trinomial

mrs_skeptik [129]

No it is not the correct

4 0

3 years ago

A garden has a length of 90 feet and a width of 60 feet. what is the area of the garden in square yards? use 1 yard = 3 feet.

Sunny_sXe [5.5K]

You just need to multiply
$90 \times 60 \div 3$
then divide by 3

7 0

3 years ago

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

Ber [7]

Given that $y_1=e^{2x/3}$ , we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$ .

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$ , so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

7 0

3 years ago

Hurry need help now 16 points!

DanielleElmas [232]

10) 8 x 5 x 10.4 = 416mm

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4 years ago