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2 years ago

I need to know exactly how to write "find the equation of the line below"

Mathematics

2 answers:

MAVERICK [17]2 years ago

8 0

Answer:

y = -4/5x + 14/5

Step-by-step explanation:

Point (6,-2) and (-4,6)

Slope = (6- - 2) / (-4-6) = 8 / -10 = - 4/5

(The slope is negative because La línea is going downhill. Uphill is positive)

You can also find the slope in the graph: rise/run.

y-intersect (any point)

Point (6, -2) (any point)

y-Intercept: -2 - (-4/5)(6) = -2 + 24/5 = 14/5

ExtremeBDS [4]2 years ago

4 0

Answer:

The general equation of the line is 4x+5y−14=0. the y=mx+c = <u>y = -4/5x + 14/5</u>

Step-by-step explanation:

Step 1) Write down your coordinates = ( -4, 6 ) and ( 6, -2 ) Step 2) find the gradient y2-y1/ x2 - x1 = -2-6 / 6--4 = -8/ 10 = -0.8 = - 4/5 = slope However in the point slope formula we need to use 3 identities and as (0, 2.4) is present coordinate for y intercept = (2.4 - 6) / (0- -4 )= -3.6/4 = - 0.9 and (-2 -2.4)/ (6-0) = -4.4/ 6 = - 0.733 we know y it crosses at 2.8 however it recognises it is a descending value (the paramount difference at 2nd point = -0.9 -0.7333= -0.1667 and widest point 0.2667 - 0.1667 = -1 would be our ascending parallel multiplier

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Answer:

1) Option A) is correct

The given rational exponent expression is not simplified correctly as a radical expression is

$x^{\frac{7}{4}}=\sqrt[7]{x^4}$

2)Option A) is correct

That is $(729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}$

Step-by-step explanation:

1) Given that $x^{\frac{7}{4}}=(\sqrt{x^7})^\frac{1}{4}$

$x^{\frac{7}{4}}=\sqrt[4]{x^7}$ is the correct answer but in the given problem they gave the RHS as wrong.

Therefore the given rational exponent expression is not simplified correctly as a radical expression is

$x^{\frac{7}{4}}=\sqrt[7]{x^4}$

2)Given that the rational exponent expression is $(729x^3y^{-18})^{\frac{1}{6}}$

To find it as a radical expression:

$(729x^3y^{-18})^{\frac{1}{6}}=(729)^{\frac{1}{6}}(x^3)^{\frac{1}{6}}((y^{-18})^{\frac{1}{6}})$

$=3(x^{\frac{3}{6}})(y^{\frac{-18}{6}})$

$=3(x^{\frac{1}{2}})(y^{-3})$

$=\frac{3\sqrt{x}}{y^3}$

Therefore $(729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}$

Therefore Option A) is correct

That is $(729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}$

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