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We have that
<span>triangle ABC
where
A(-5, 5), B(1, 1), and C(3, 4) are the vertices
using a graph tool
see the attached figure
the hypotenuse is the segment AC
find the equation of the line AC
</span>A(-5, 5) C(3, 4)
<span>
step 1
find the slope m
m=(y2-y1)/(x2-x1)-----> m=</span>(4-5)/(3+5)-----> m=-1/8
step 2
with C(3,4) and m=-1/8
find the equation of a line
y-y1=m*(x-x1)-----> y-4=(-1/8)*(x-3)----> y=(-1/8)*x+(3/8)+4
y=(-1/8)*x+(3/8)+4----> multiply by 8----> 8y=-x+3+32
8y=-x+35
the standard form is Ax+By=C
so
x+8y=35
A=1
B=8
C=35
the answer is
x+8y=35
<span>triangle ABC
where
A(-5, 5), B(1, 1), and C(3, 4) are the vertices
using a graph tool
see the attached figure
the hypotenuse is the segment AC
find the equation of the line AC
</span>A(-5, 5) C(3, 4)
<span>
step 1
find the slope m
m=(y2-y1)/(x2-x1)-----> m=</span>(4-5)/(3+5)-----> m=-1/8
step 2
with C(3,4) and m=-1/8
find the equation of a line
y-y1=m*(x-x1)-----> y-4=(-1/8)*(x-3)----> y=(-1/8)*x+(3/8)+4
y=(-1/8)*x+(3/8)+4----> multiply by 8----> 8y=-x+3+32
8y=-x+35
the standard form is Ax+By=C
so
x+8y=35
A=1
B=8
C=35
the answer is
x+8y=35
Observe the given data distribution table carefully.
The 5th class interval is given as,
The upper limit (UL) and lower limit (LL) of this interval are,
Thus, the upper-class limit of this 5th class is 17.4.