For the case of a line through
(
2
,
7
)
and
(
1
,
−
4
)
we have
slope
m
=
Δ
y
Δ
x
=
−
4
−
7
1
−
2
=
−
11
−
1
=
11
Using the point
(
2
,
7
)
we can write the equation of the line in point slope form as:
y
−
7
=
m
(
x
−
2
)
where the slope
m
=
11
.
That is:
y
−
7
=
11
(
x
−
2
)
To get point intercept form, first expand the right hand side so...
y
−
7
=
11
x
−
(
11
⋅
2
)
=
11
x
−
22
Then add
7
to both sides to get:
y
=
11
x
−
15
=
11
x
+
(
−
15
)
This is point intercept (
y
=
m
x
+
c
) form with slope
m
=
11
and intercept
c
=
−
15
.
Answer:
10sqrt3+22
Step-by-step explanation:
Ok, let us imagine it as a sort of rectangle split upon its diagonal.
Using that, we can Pythag it out,
11^2+b^2=14^2
121+b^2=196
b^2=75
b=sqrt75
b=5sqrt3
Ok, using this info, we find the perimeter,
5sqrt3+5sqrt3+11+11
10sqrt3+22
The answer is 10sqrt3+22
<span>B. It must be the same as when he constructed the arc centered at point A.
This problem would be a lot easier if you had actually supplied the diagram with the "arcs shown". But thankfully, with a few assumptions, the solution can be determined.
Usually when constructing a perpendicular to a line through a specified point, you first use a compass centered on the point to strike a couple of arcs on the line on both sides of the point, so that you define two points that are equal distance from the desired intersection point for the perpendicular. Then you increase the radius of the compass and using that setting, construct an arc above the line passing through the area that the perpendicular will go. And you repeat that using the same compass settings on the second arc constructed. This will define a point such that you'll create two right triangles that are reflections of each other. With that in mind, let's look closely at your problem to deduce the information that's missing.
"... places his compass on point B ..."
Since he's not placing the compass on point Q, that would imply that the two points on the line have already been constructed and that point B is one of those 2 points. So let's look at the available choices and see what makes sense.
A .It must be wider than when he constructed the arc centered at point A.
Not good. Since this implies that the arc centered on point A has been constructed, then it's a safe assumption that points A and B are the two points defined by the initial pair of arcs constructed that intersect the line and are centered around point Q. If that's the case, then the arc centered around point B must match exactly the setting used for the arc centered on point A. So this is the wrong answer.
B It must be the same as when he constructed the arc centered at point A.
Perfect! Look at the description of creating a perpendicular at the top of this answer. This is the correct answer.
C. It must be equal to BQ.
Nope. If this were the case, the newly created arc would simply pass through point Q and never intersect the arc centered on point A. So it's wrong.
D.It must be equal to AB.
Sorta. The setting here would work IF that's also the setting used for the arc centered on A. But that's not guaranteed in the description above and as such, this is wrong.</span>
Answer: 3/5 = (6/10). There are 10 TOTAL marbles in the bag.
Step-by-step explanation:
Since it is a question of fractions, you can assume any reasonable number and work with the numbers. We know initially that there are 6 marbles and they are ALL blue and the rest are another color or colors. So BLUE = 6/10 and the other colors 4/10 for 2/5.
Answer:
Find the stated lengths and values of rectangle M 26 N
LMNO. Round to the nearest tenth. P 5a - 7 2a + 8 240 9. MO
10. LP O 11. LM 12. ZNMO Find the missing leng..
