Ask question

Ask question

All categories

3 years ago

11

2/3y = 4/5 solve the equation using inverse operations

1 answer:

Alisiya [41]3 years ago

6 0

Answer:

y=6/5

Step-by-step explanation:

Hope this helps! Sorry if its wrong

You might be interested in

The population of small town grew from 4732 to 5967 from

Svetlanka [38]

Answer:

the small town grew to 2022

4 0

2 years ago

A^3+a+a^2+1 2a^2+2ab+ab^2+b^3

Vikentia [17]

$\frac{a^3+a+a^2+1}{a^3+a^2+ab^2+b^2} \cdot \frac{2a^2+2ab+ab^2+b^3}{a^3+a+a^2b+b} =\frac{a(a^2+1)+1(a^2+1)}{a^2(a+1)+b^2(a+1)} \cdot \frac{2a(a+b)+b^2(a+b)}{a(a^2+1)+b(a^2+1)} =\\\\=\frac{(a^2+1)(a+1)}{(a+1)(a^2+b^2)} \cdot \frac{(a+b)(2a+b^2)}{(a^2+1)(a+b)} = \frac{2a+b^2}{a^2+b^2}$

6 0

3 years ago

1 quart=4 cups (whats the answer?)

Sever21 [200]

Answer:

1quart 4cups =8quatcups

Step-by-step explanation:

thanks:)

7 0

2 years ago

Read 2 more answers

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in <em>c)</em>

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

Pls help me guys!<br> Thanks :D Pls help me I'm stuck

MariettaO [177]

C. 21,525 < 22,789 hope this helps

7 0

3 years ago

Read 2 more answers