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2 years ago

15

Hi please help me with this i’ll give brainliest if you give a correct and show your work!

1 answer:

Tatiana [17]2 years ago

6 0

1. (1.35)x(.42)

1.35
x. .42
________
270
5400
________
.5670

You need to have the total number of decimal places from the original two numbers. 1.35 has two decimal places and .42 has to decimal places you have a total of four decimal places, Which means 5670 is .5670

2. (.22)x(.04)

.22
x. .04
_______
88
00
_______
.0088

you need to have the total number of decimal places from the original Two numbers. So .22 is two decimal places and .04 is two decimal places, so you will need to have for decimal places. That means that the answer 88 needs to be translated into .0088

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3 years ago

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Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0

natita [175]

Answer:

$\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

Step-by-step explanation:

Given differential equation,

$(x^2 + y^2) dx + (x^2 - xy) dy = 0$

$\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)$

Let y = vx

Differentiating with respect to x,

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

From equation (1),

$v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}$

$v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}$

$v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}$

$v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}$

$x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v$

$x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}$

$x\frac{dv}{dx}=\frac{v+1}{v-1}$

$\frac{v-1}{v+1}dv=\frac{1}{x}dx$

Integrating both sides,

$\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx$

$\int{\frac{v-1+1-1}{v+1}}dv=lnx + C$

$\int{1-\frac{2}{v+1}}dv=lnx + C$

$v-2ln(v+1)=lnx+C$

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$\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

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