Determine whether the two lines are parallel, perpendicular or neither x-3y=6 y=3x+4

Mathematics

1 answer:

Anna [14]4 years ago

6 0

Answer:

neither

Here's how you know:

first, i'll simplify it so both are in slope-intercept form (y = mx + b).

x-3y=6

-3y = 6 - x

y = (1/3)x - 2

y=3x+4 is already in slope-intercept form, yay :)

these lines are neither parallel nor perpendicular.

The equations of parallel lines have THE SAME SLOPES (y = mx + b --> m would be exactly the same).

In contrast, perpendicular lines have NEGATIVE RECIPROCAL SLOPES (y = mx + b --> m would be -1/m).

--> since the slopes of these match neither of these definitions, we know it must be neither.

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PLEASE HELP!!!!

Lisa [10]

Answer:

x = - 3 with multiplicity 2

Step-by-step explanation:

f(x) = (x - 3)(x + 3)(x + 3) = (x - 3)(x + 3)²

Equating each factor to zero and solving for x

x - 3 = 0 ⇒ x = 3 with multiplicity 1

x + 3 = 0 ⇒ x = - 3

x + 3 = 0 ⇒ x = -3

Thus x = - 3 has multiplicity of 2

The fact that the factor is squared gives the multiplicity

(x + 3)³ has root - 3 of multiplicity 3

8 0

3 years ago

How do you find the area of this composite figure?

Illusion [34]

If you notice the picture below

the composite figure is just a trapezoid sitting on top of a rectangle
and then, the rectangle has a triangular hole in it

so.. get the area of the trapezoid $\bf \textit{area of a trapezoid}=A=\cfrac{h}{2}(a+b)\qquad 
\begin{cases}
h=height\\
a,b=\textit{parallel sides or bases}
\end{cases}$

then get the area of the rectangle, which is just a 12x14
and then get the area of the triangle, which surely you know is 1/2 bh

then, subtract the triangle's area from the rectangle's area

and whatever is left, namely the difference, add that to the area of the trapezoid, and that's the composite's area

namely the area of the trapezoid plus the rectangle's, minus the triangle's