Math 2 algebra questions
Assuming the 'standard form' is "slope-intercept", calculate the slope from the equation m = y2-y1 = 5 - 1 = 4 = -2
x2-x1 4 - 6 -2
The intercept can be found by substituting either of the two points into the equation y = mx + b
5 = (-2)4 + b
5 = (-8) + b
13 = b
(OR, using the other point, y = mx + b
1 = (-2)6 + b
1 = (-12) + b
13 = b )
Then expressing in general:
y = (-2) x + 13
SOURCE: math
Divide the first equation by 2, the second by 6 (i.e. the number that comes before the x)
SOURCE: write an equation of the line in standard form that pass through (-5,-11) and 10,7)
Calcualte the slope of the line as
a=(7-(-11))/(10-(-5))=18/15=6/5
Use the fact that the line passes through one of the two points, for example (10,7)
7=(6/5)*10+b=12+b
Obtain b as b=7-12=-5
The equation of the line in functional form is y=(6/5)x-5
Multiply everything by 5 to clear the fraction
5y=6x-25 or 0=6x-5y-25
Finally, the equation in general form (standard?) is 6x-5y-25=0.
Check the calculation by verifying that the point (10,7) lies on the line.
6(10)-5(7)-25=60-35-25=60-60=0 CHECKed!
Check that the second point (-5,-11) lies on the line also (if you want to)
6*(-5)-5*(-11)-25=-30+55-25=0
That checks OK.
SOURCE: write each quadratic equation in standard form. .
5X-10X^2=12
0=10X^2-5X+12
Standard form is aX^2+bX+c=0 <===> 10X^2+(-5)X+12=0
Identifying the factors of X^2 gives a=10
Identifying the factors of X gives b=-5
Identifying the constant terms gives c=12
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