Determine the standard form of the equation of the line that passes through (6, -9) and (-8, 6)
1 answer:
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Answer:
A. 15x +14y = -36
Step-by-step explanation:
Since we are given two points, we can start with the 2-point form of the equation for a line.
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
y = (6 -(-9))/(-8 -6)(x -6) +(-9)
y = 15/-14(x -6) -9
Multiplying by -14, we have ...
-14y = 15x -90 +126
Adding 14y-36 to both sides gives ...
-36 = 15x +14y . . . . matches choice A
The standard-form equation is ...
15x +14y = -36
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<em>Additional comments</em>
It can be easier to start with the form ...
(Δy)x -(Δx)y = (Δy)x1 -(Δx)y1 . . . . . where Δx = x2-x1 and Δy = y2-y1
This gives ...
(6+9)x -(-8-6)y = 15(6) +14(-9)
15x +14y = -36 . . . simplified
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You can also start with the slope-intercept form or the point-slope form, if you're more familiar with those. The result will be the same. I find it handy to be familiar with a number of different forms of the equation for a line.
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Answer:
The probability that you will get a sum of 4 before you get a sum of 5 is
Step-by-step explanation:
Given : If you roll two fair dice repeatedly.
To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?
Solution :
When two dice are rolled the outcomes are
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Total number of outcomes = 36
Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3
The probability that you will get a sum of 4 before you get a sum of 5 is
Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is