How do you find an equation for a line with a slope of 2/3 passing through (1,-4)

Mathematics

2 answers:

Igoryamba3 years ago

7 0

Answer:

The standard general equation for a straight line is:

y = mx + C. You said the slope of line is 2/3. m is always the slope in this context. So with the numerical value slope we have the line

Y = 2/3 X + C. You also said the line passes through the point (1, -4) that is the X value is 1 and the y value is -4. Substitute these value into the linear equation to find C, as follows

-4 = 2/3 + c. and so C = - 4 - 2/3 = -14/3 and so the equation of line is :

y = 2x/3 + 14/3 or if you like 3y = 2x + 14

It is important to remember that the convention of the bracketted numbers e.g. (1,-4) or ( 1, -4, z), say is that the first number in the brackets indicates the X value, the second the y value and the third, the z value in the case of 3 D coordinates.

mart [117]3 years ago

3 0

*Y=2/3x-6
I believe this is the answer!!! Hope you get a good grade

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andreev551 [17]

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WHAT IS X³-27 SIMPLIFIED

Eduardwww [97]

Answer:

It is (x - 3)³ - 9x(3 - x)

Step-by-step explanation:

Express 27 in terms of cubes, 27 = 3³:

$= {x}^{3} - {3}^{3}$

From trinomial expansion:

${(x - y)}^{3} = (x - y)(x - y)(x - y) \\$

open first two brackets to get a quadratic equation:

${(x - y)}^{3} = ( {x}^{2} - 2xy + {y}^{2} )(x - y)$

expand further:

${(x - y)}^{3} = {x}^{3} - y {x}^{2} - 2y {x}^{2} + 2x {y}^{2} + x {y}^{2} - {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3x {y}^{2} - 3y {x}^{2} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3xy(y - x) \\ \\ { \boxed{( {x}^{3} - {y}^{3} ) = {(x - y)}^{3} - 3xy(y - x)}}$