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3 years ago

What point is on both lines y = {x +3 and y = x +1? (1 point) Please help ASAP

Mathematics

1 answer:

Stolb23 [73]3 years ago

3 0

Answer:

No point

Step-by-step explanation:

This is a trick question because if you break down the graphing equations to y=mx+b, you see that it is a linear graph, so it will have a straight line. For the first equations, y=x+3, you look at the original equations, y=mx+b. m=slope, b=x-intercept. Slope is rise/run (y/x) so the slope would be up one, over one (1/1=1). The x-intercept is 3 so its a straight line but it intercepts at 3. The next equation is the exact same thing, but the x-intercept is at 1. A straight line that intercepts at 1. So it is two parallel lines that never ever touch. So this is a trick question.

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nevsk [136]

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3 years ago

bija089 [108]

Answer:

The Parenthesis tells you what operation to do first.

Step-by-step explanation:

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4 0

3 years ago

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,

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

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

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

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

Now, y = vx ⇒ v = y/x

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

5 0

3 years ago

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lbvjy [14]

Answer:

Step-by-step explanation:

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