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3 years ago

The slope of the line below is -3. Use the coordinates of the labeled point to find a point-slope equation of the line.

1 answer:

8 0

Answer:
y-y₁ = -3(x-x₁)

First of all, when you start an equation with a slope, since the question gave the slope -3, I'm not sure what your looking for, but I'll give you an example

Example:
y-y₁ = m(x-x₁)
m = -3
y₁ = ? - I'll just put a random # = -1
x₁ = ? - I'll just put a random # = 5
<span>y-(-1) = (-3)(x-(5))
</span>y+1 = -3x+15 - I'm guessing

You might be interested in

−2(3x−4)−x−3=<br> \,\,33<br> 33

notsponge [240]

Answer: $=-7x+5$

Step-by-step explanation:

$-2\left(3x-4\right)-x-3$

$=-\left(2\left(3x-4\right)+x+3\right)$

$=-\left(6x-8+x+3\right)$

$=-\left(7x-5\right)$

$=-7x+5$

5 0

3 years ago

Find the inverse of the given function.

kodGreya [7K]

For this case we must find the inverse of the following function:

$f (x) = - \frac {1} {2} \sqrt {x + 3}$

We follow the steps below:

Replace f(x) with y:

$y = -\frac {1} {2} \sqrt {x + 3}$

We exchange the variables:

$x = - \frac {1} {2} \sqrt {y + 3}$

We solve for "y":

$- \frac {1} {2} \sqrt {y + 3} = x$

Multiply by -2 on both sides of the equation:

$\sqrt {y + 3} = - 2x$

We raise both sides of the equation to the square to eliminate the radical:

$(\sqrt {y + 3}) ^ 2 = (- 2x) ^ 2\\y + 3 = 4x ^ 2$

We subtract 3 from both sides of the equation:

$y = 4x ^ 2-3$

We change y by f ^ {- 1} (x):

$f ^ {- 1} (x) = 4x ^ 2-3$

Answer: $f ^ {- 1} (x) = 4x ^ 2-3$

7 0

3 years ago

Read 2 more answers

Which equation is the direct variation equation if f(x) varies directly with x and f(x)=-18 when x=3

Otrada [13]

$\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{f(x)}{y}=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
\textit{we also know that }
\begin{cases}
y=-18\\
x=3
\end{cases}\implies -18=k3
\\\\\\
\cfrac{-18}{3}=k\implies -6=k\qquad therefore\qquad \boxed{\stackrel{f(x)}{y}=-6x}$