Write the equation of the line that passes through (3, 4) and (2, −1) in slope-intercept form.

1 answer:

3 0

(3,4)(2,-1)
slope(m) = (y2 - y1) / (x2 - x1)
slope(m) = (-1-4) / (2 - 3) = -5/-1 = 5

y = mx + b
slope(m) = 5
u can use either of ur points...(3,4)...x = 3 and y = 4
now we sub and find b, the y int
4 = 5(3) + b
4 = 15 + b
4 - 15 = b
- 11 = b

so ur equation is : y = 5x - 11 <===

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Write an equation of the line that passes through the given point and is parallel to the graph of the given equation.

True [87]

Question 1:

--------------------------------------------------------------------
Find Slope
--------------------------------------------------------------------
Equation: y = 5x - 2
Slope = 5
Slope of parallel line = 5

--------------------------------------------------------------------
Insert slope into the general equation y = mx + c
--------------------------------------------------------------------
y = 5x + c

--------------------------------------------------------------------
Find y-intercept
--------------------------------------------------------------------
At point (2, -1)
y = 5x + c
-1 = 5(2) + c
c = -1 - 10
c = -11

--------------------------------------------------------------------
Insert y-intercept into the equation
--------------------------------------------------------------------
y = 5x + c
y = 5x - 11

--------------------------------------------------------------------
Answer: y = 5x - 11
--------------------------------------------------------------------

Question 2:

--------------------------------------------------------------------
Find Slope
--------------------------------------------------------------------
y = 9x
Slope = 9
Slope of the parallel line = 9

--------------------------------------------------------------------
Insert slope into the equation y = mx + c
--------------------------------------------------------------------
y = 9x + c

--------------------------------------------------------------------
Find y-intercept
--------------------------------------------------------------------
y = 9x + c
At point (0, 5)
5 = 9(0) + c
c = 5

--------------------------------------------------------------------
Insert y-intercept into the equation
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y = 9x + c
y = 9x + 5

--------------------------------------------------------------------
Answer: y = 9x + 5
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13. Given the following historical data and weights of .5, .3, and .2, what is the three-period weighted moving average forecast

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Answer:

b. 144.8

Step-by-step explanation:

When calculating the moving average estimate of an observation , each of the observations are usually computed with the same weighted . In some cases, it is beneficial to assign different weight on the observations such that the observation closer to the time period being forecast, has higher weight. This is refer to as weighted moving average technique. The sum of the individual weight in a weighted moving average technique must equal to 1.

The three-period weighted moving average forecast for period 5 = 144*0.5 + 148 *0.3 + 142 *0.2 = 144.8

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Answer:

20 cm

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Length of poles : 8cm ; h

8cm long pole = shadow length, 6m

h cm long pole = shadow length, 15 m

Using cross multiplication :

8cm * 15 m = h cm * 6m

120 = 6h

h = 120 / 6

h = 20 cm

Height of second pole = 20cm

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