Q: What is the answer to yx-5 if y is 0?

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y = x^5 + x^3 + x^2 + x y' = 5x^(5 - 1) + 3x^(3 - 1) + 2x^(2 -1) + 1x^(1 - 1) (note x^1 = x and x^0 = 1) y' = 5x^4 + 3x^2 + 2x + 1

If you mean: y = -15x+60 then 15x+y = 60 When y = 0 then x = 4 When x = 0 then y = 60 Coordinates of the line are: (4, 0) and (0, 60)

4

y = 2x + 5At the y-intercept, x=0.y = 0 + 5y = 5

You state that y + x = -8 The y intercept is where x = 0 so replace that in the formula: y + 0 = -8 So the y intercept is at -8 ... (0, -8) is the point you're looking for.

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Put simply, the inverse of y=x^5+x^4+x is x=y^5+y^4+y. Unfortunately, this is a quintic function and there is no quintic formula.

y = x^5 + x^3 + x^2 + x y' = 5x^(5 - 1) + 3x^(3 - 1) + 2x^(2 -1) + 1x^(1 - 1) (note x^1 = x and x^0 = 1) y' = 5x^4 + 3x^2 + 2x + 1

x-y=0 x=y so you will use that in the other equation by substituting every y with x 2x+y=0 2x+x=0 3x=0 x=0/3 x=0 then use that in the previous equation by substituting every x with 0 x=y 0=y; y=0 finally x=0 and y=0

In 2-d: (0, y) In 3-d: (0, y, 0) In 4-d: (0, y, 0, 0) and so on.

3x + 2y = x + y y = -2x to find y-int set x equal to zero y = -2(0) = 0 or... 3(0) + 2y = (0) + y 2y - y = 0 y = 0 y-intercept is zero

Solve y''+y=0 using Laplace. Umm y=0, 0''+0=0, 0.o Oh well here it is. First you take the Laplace of each term, so . . . L(y'')+L(y)=L(0) Using your Laplace table you know the Laplace of all these terms s2L(y)-sy(0)-y'(0) + L(y) = 0 Since both initial conditions are 0 this simplifies to. . . s2L(y) + L(y) = 0 You can factor out the L(y) and solve for it. L(y) = 0/(s2+1) L(y) = 0 Now take the inverse Laplace of both sides and solve for y. L-1(L(y)) = L-1(0) y = 0

The y-intercept of y = x is 0. (At the point (0,0) )

16y-9y3 = 0 y*(16-9y2) = 0 y*(4-3y)*(4+3y) = 0 So y = 0 or (4-3y) = 0 or (4+3y) = 0 ie y = 0 or y = 4/3 or y = -4/3

y intercept: x = 0 3y = 0 y = 0 x intercept: y=0 -4x = 0 x = 0 (0,0)

2+y-1=0 i.e. y+1=0 Hence, y= -1.

Yes, they can both be zero.

If y = ax and a = 0, then y=0. No matter what x is, y is still 0. Therefore any graph of y=ax (where a=0) will simply be a line at y=0, which is the x-axis.