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borishaifa [10]

3 years ago

10

Suppose you are going to graph the data in the table below. What data should be represented on each axis, and what should the in

crements be? Hours Distance traveled (miles) 1 60 2 35 3 45 4 65 5 55 6 62 7 12 8 65

2 answers:

Brrunno [24]3 years ago

8 0

Distance traveled should be on y axis
hours should be on x axis
increments should be 5

Ugo [173]3 years ago

7 0

Answer:

The data is given in the form of table as :

Hours Distance traveled in miles

1 60

2 35

3 45

4 65

5 55

6 62

7 12

8 65

Since, the values of hours in numerically smaller than than the values of distance traveled in miles so we take the value of hours on x-axis and the distance traveled in miles on y-axis

For, The increment : On the x-axis we take the increment of 5 units and on the y-axis we take the increment of 10 units

And the resultant graph is attached below :

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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),

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

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

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

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

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

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

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

Integrating both sides,

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

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

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

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

Now, y = vx ⇒ v = y/x

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

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3 years ago