- 100 points -
2 answers:
D) - y = (x + 5)(x - 3)(x + 1).
<h3>EXPLANATION:</h3>Table in this case would look like this:
Write coefficients of x³, x² ,x and the constant in a row and divisor would be the x value obtained by equation x + 5 = 0.
The sequence of multiplications would be as shown in picture.
x² - 2x - 3
x² - 3x + x - 3
x(x - 3) + 1(x - 3)
(x + 1)(x - 3)(x + 5).
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I attached the work and answer for this problem. If you have any questions please feel free to ask.
Hope this helps.
Let
A (x1,y1)-------> (1,5)
B (x2,y2)------> (5,10)
C (x3,y3)------> (9,15)
D (x4,y4)------> (13,20)
find the slope
AB
m=(10-5)/(5-1)-----> m=5/4
find the slope
CB
m=(15-10)/(9-5)-----> m=5/4
find the slope
DC
m=(20-15)/(13-9)-----> m=5/4
so
<span>the relationship shown is linear
</span>
<span>to model the data with an equation find the equation of a line
</span>with m=5/4 and point A (1,5)
y-y1=m*(x-x1)-----> y-5=(5/4)*(x-1)----> y=1.25x-1.25+5-----> y=1.25x+3.75
the equation to model data is
y=1.25x+3.75
see the attached figure
A (x1,y1)-------> (1,5)
B (x2,y2)------> (5,10)
C (x3,y3)------> (9,15)
D (x4,y4)------> (13,20)
find the slope
AB
m=(10-5)/(5-1)-----> m=5/4
find the slope
CB
m=(15-10)/(9-5)-----> m=5/4
find the slope
DC
m=(20-15)/(13-9)-----> m=5/4
so
<span>the relationship shown is linear
</span>
<span>to model the data with an equation find the equation of a line
</span>with m=5/4 and point A (1,5)
y-y1=m*(x-x1)-----> y-5=(5/4)*(x-1)----> y=1.25x-1.25+5-----> y=1.25x+3.75
the equation to model data is
y=1.25x+3.75
see the attached figure