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3 years ago

A student finds the slop of the line between (4, 3) and (-2, -2). She writes -2–4/-2–3. What mistake did she make?

2 answers:

7 0

Her mistake is that she did the change in x over the change in y. It should be the change in y over the change in x. So: -2-3/-2-4

Aleonysh [2.5K]3 years ago

3 0

In the slope formula, The difference in the y-values are on the numerator and the difference in the x-values are on the denominator.
It should be:
(-2-3)/(-2-4)

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Find the value of the variable y, where the sum of the fractions 6/(y+1) and y/(y-2) is equal to their product.

Blizzard [7]

Answer:

The answer is

$y = 3$

$y = - 4$

Step-by-step explanation:

We must find a solution where

$\frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Consider the Left Side:

First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get

$\frac{6}{y + 1} \times \frac{y - 2}{y - 2} + \frac{y}{y - 2} \times \frac{y + 1}{y + 1}$

Which equals

$\frac{6y - 12}{(y -2) (y + 1)} + \frac{ {y}^{2} + y }{(y - 2)(y + 1)}$

Add the fractions

$\frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Simplify the right side by multiplying the fraction

$\frac{6y}{(y + 1)(y + 2)}$

Set both fractions equal to each other

$\frac{6y}{(y + 1)(y - 2)} = \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)}$

Since the denomiator are equal, we must set the numerator equal to each other

$6y = {y}^{2} + 7y - 12$

$= {y}^{2} + y - 12$

$(y + 4)(y - 3)$

$y = - 4$

$y = 3$

6 0

3 years ago

Read 2 more answers

Rewrite the equation 2x + 4y + 2 = 3y + 5 in general form. A. 2x + 7y + 7 = 0 B. 2x + y – 3 = 0 C. 2x + y = 3 D. y = –2x + 3

kvasek [131]

2x + 4y + 2 = 3y + 5
2x + 4y - 3y + 2 - 5 = 0
2x + y - 3 = 0

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]

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