Find the solution for this 3y = 6x - 3 y = 2x - 1

Mathematics

1 answer:

kondaur [170]3 years ago

8 0

There are infinitely many solutions because the two equations are the same line.

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torisob [31]

Answer:

first one with -1...21

Step-by-step explanation:

only one that follows quadratic formula format

5 0

3 years ago

What is the equation of the line?

prohojiy [21]

Answer:

A) y = -1/2x + 1/2

Step-by-step explanation:

Find the y-intercept(when x = 0), which is 1/2.

Find the slope: m = y2-y1 / x2-x1

I used points: (3, -1), (-3,2)

m = 2 - (-1) / -3 - 3

m = 3 / - 6

m = -1/2

plug this into the slope intercept form equation: y = mx + b

y = -1/2x + 1/2

7 0

3 years ago

The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.