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3 years ago

Given points A (-5, 6) and B (7, -1), find: Distance, Midpoint, and Slope

1 answer:

4 0

We have that
points A (-5, 6) and B (7, -1)

Part A)
find the distance
d=√[(y2-y1)²+(x2-x1)²]-------> d=√[(-1-6)²+(7+5)²]----> d=√(49+144)
d=√193 units

Part B)
find the midpoint
ABx=(x1+x2)/2-----> (-5+7)/2-----> ABx=1
ABy=(y1+y2)/2-----> (-1+6)/2-----> ABy=2.5

the midpoint is (1,2.5)

Part C)
find the slope
m=(y2-y1)/(x2-x1)-----> m=(-1-6)/(7+5)--------> m=-7/12

the slope m=-7/12

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4) What is the equation of the line that passes through the points (3,0) and 2 points

liubo4ka [24]

Presume y=kx +b,then substitute (3,0)&(5,4),
When x=3,y=0,so 3k+b=0;
When x=5,y=4,so 5k+b=4;
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The answer is:y=2x-6.

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3 years ago

Find K by evaluating Limit

Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

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{L'Hospital's rule}

I hope this helps you

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6 0

1 year ago

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2 years ago

Use substitution to solve the following system of equations. What is the value of y? {3x+2y=12{5x−y=7 A) y = -3B) y = 3C) y = -2

Vlad [161]

Answer

B) y = 3

Step-by-step explanation

Given the system of equations:

$\begin{gathered} 3x+2y=12\text{ \lparen eq. 1\rparen} \\ 5x-y=7\text{ \lparen eq. 2\rparen} \end{gathered}$

Isolating x from equation 1:

$\begin{gathered} 3x+2y-2y=12-2y \\ 3x=12-2y \\ \frac{3x}{3}=\frac{12-2y}{3} \\ x=\frac{12}{3}-\frac{2}{3}y \\ x=4-\frac{2}{3}y\text{ \lparen eq. 3\rparen} \end{gathered}$

Substituting equation 3 into equation 2 and solving for y:

$\begin{gathered} 5(4-\frac{2}{3}y)-y=7 \\ 5\cdot4-5\cdot\frac{2}{3}y-y=7 \\ 20-\frac{10}{3}y-y=7 \\ 20-\frac{13}{3}y=7 \\ 20-\frac{13}{3}y-20=7-20 \\ -\frac{13}{3}y=-13 \\ (-\frac{3}{13})\cdot-\frac{13}{3}y=(-\frac{3}{13})\cdot-13 \\ y=3 \end{gathered}$

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10 months ago