The expression above can also be written in the form of
1 answer:
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Let
<span>A------> (−3, −11)
B------> (6, 4)
step 1
find the slope m point A and point B
m=(y2-y1)/(x2-x1)-----> </span>m=(4+11)/(6+3)----> m=15/9---> 5/3
step 2
with m=5/3 and point B (6, 4) find the equation of a line
y-y1=m*(x-x1)-----> y-4=(5/3)*(x-6)---> y=(5/3)x-6
case <span>a. they have the same slope------> is not correct
because
</span><span>y=35x−6------> the slope m=35
</span>y=(5/3)x-6---> the slope m =5/3
case <span>b. they have the same x-intercept.----> is not correct
because
</span>the x-intercept is for y=0
y=35x−6-----> 0=35x-6----> 35x=6-----> x=6/35
the x-intercept is the point (6/35,0)
y=(5/3)x-6----> 0=(5/3)x-6----> (5/3)x=6---> x=18/5
the x-intercept is the point (18/5,0)
case <span>c. the two lines are perpendicular. do. they have the same y-intercept-----> is nor correct
because
if two lines are perpendicular
m1*m2=-1-------> </span><span>this condition is not satisfied
</span>
the y intercept is for x=0
y=35x−6-------> y=-6
the y intercept is the point (0,-6)
y=(5/3)x-6-----> y=-6
the y intercept is the point (0,-6)
they have the same y-intercept but the two lines are not perpendicular
see the attached figure
<span>A------> (−3, −11)
B------> (6, 4)
step 1
find the slope m point A and point B
m=(y2-y1)/(x2-x1)-----> </span>m=(4+11)/(6+3)----> m=15/9---> 5/3
step 2
with m=5/3 and point B (6, 4) find the equation of a line
y-y1=m*(x-x1)-----> y-4=(5/3)*(x-6)---> y=(5/3)x-6
case <span>a. they have the same slope------> is not correct
because
</span><span>y=35x−6------> the slope m=35
</span>y=(5/3)x-6---> the slope m =5/3
case <span>b. they have the same x-intercept.----> is not correct
because
</span>the x-intercept is for y=0
y=35x−6-----> 0=35x-6----> 35x=6-----> x=6/35
the x-intercept is the point (6/35,0)
y=(5/3)x-6----> 0=(5/3)x-6----> (5/3)x=6---> x=18/5
the x-intercept is the point (18/5,0)
case <span>c. the two lines are perpendicular. do. they have the same y-intercept-----> is nor correct
because
if two lines are perpendicular
m1*m2=-1-------> </span><span>this condition is not satisfied
</span>
the y intercept is for x=0
y=35x−6-------> y=-6
the y intercept is the point (0,-6)
y=(5/3)x-6-----> y=-6
the y intercept is the point (0,-6)
they have the same y-intercept but the two lines are not perpendicular
see the attached figure
Answer:
16 students can sit around a cluster of 7 square table.
Step-by-step explanation:
Consider the provided information.
We need to find how many students can sit around a cluster of 7 square table.
The tables in a classroom have square tops.
Four students can comfortably sit at each table with ample working space.
If we put the tables together in cluster it will look as shown in figure.
From the pattern we can observe that:
Number of square table in each cluster Total number of students
1 4
2 6
3 8
4 10
5 12
6 14
7 16
Hence, 16 students can sit around a cluster of 7 square table.
