Whts the solution from this?
1 answer:
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Answer:
* The equation of the median of the trapezoid is 10x + 6y = 39
Step-by-step explanation:
* Lets explain how to solve the problem
- The slope of the line whose end points are (x1 , y1) , (x2 , y2) is
- The mid point of the line whose end point are (x1 , y1) , (x2 , y2) is
- The standard form of the linear equation is Ax + BC = C, where
A , B , C are integers and A , B ≠ 0
- The median of a trapezoid is a segment that joins the midpoints of
the nonparallel sides
- It has two properties:
# It is parallel to both bases
# Its length equals half the sum of the base lengths
* Lets solve the problem
- The trapezoid has vertices R (-1 , 5) , S (! , 8) , T (7 , -2) , U (2 , 0)
- Lets find the slope of the 4 sides two find which of them are the
parallel bases and which of them are the non-parallel bases
# The side RS
∵
# The side ST
∵
# The side TU
∵
# The side UR
∵
∵ The slope of ST = the slop UR
∴ ST// UR
∴ The parallel bases are ST and UR
∴ The nonparallel sides are RS and TU
- Lets find the midpoint of RS and TU to find the equation of the
median of the trapezoid
∵ The median of a trapezoid is a segment that joins the midpoints of
the nonparallel sides
∵ The midpoint of RS =
∵ The median is parallel to both bases
∴ The slope of the median equal the slopes of the parallel bases = -5/3
∵ The form of the equation of a line is y = mx + c
∴ The equation of the median is y = -5/3 x + c
- To find c substitute x , y in the equation by the coordinates of the
midpoint of RS
∵ The mid point of Rs is (0 , 13/2)
∴ 13/2 = -5/3 (0) + c
∴ 13/2 = c
∴ The equation of the median is y = -5/3 x + 13/2
- Multiply the two sides by 6 to cancel the denominator
∴ The equation of the median is 6y = -10x + 39
- Add 10x to both sides
∴ The equation of the median is 10x + 6y = 39
* The equation of the median of the trapezoid is 10x + 6y = 39
Problem 61
The nth triangular number is
T(n) = n(n+1)/2
I'll rewrite this into
T(n) = 0.5n(n+1)
The triangular number right after this is
T(n+1) = 0.5(n+1)(n+2)
I replaced every n with n+1 and simplified
Let's see what we get when we add up the two expressions
T(n) + T(n+1)
0.5n(n+1) + 0.5(n+1)(n+2)
0.5n^2+0.5n + 0.5(n^2+3n+2)
0.5n^2 + 0.5n + 0.5n^2 + 1.5n + 1
n^2+2n+1
(n+1)^2
This shows that the sum of any two consecutive triangular numbers results in a square number
Here's a few examples
- 0+1 = 1
- 1+3 = 4
- 3+6 = 9
- 6+10 = 16
- 10+15 = 25
Note each sum is a perfect square, which visually would plot out a square figure.
For quick reference, the set of the first few triangular numbers is {0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55,...}
<h3>Answer: Square number</h3>==========================================================
Problem 64
Let's say we go with n = 5.
This means,
T(n) = 0.5n(n+1)
T(n-1) = 0.5(n-1)(n-1+1)
T(n-1) = 0.5n(n-1)
T(5-1) = 0.5*5(5-1)
T(4) = 10
This says that when n = 5, the 4th triangular number is 10
Triple that result and add on n = 5
3*T(4) + n = 3*10+5 = 35
This result is beyond obvious which category of figurate number it belongs to. It's not a triangular number since it's not in the form n(n+1)/2. It's not a square number either.
Through a bit of trial and error, you should find it's a pentagonal number
Pentagonal numbers are of the form n(3n-1)/2
If you plugged n = 5 into that, it leads to 35
n(3n-1)/2 = 5*(3*5-1)/2 = 5*14/2 = 70/2 = 35
The diagram shown below represents the first few pentagonal numbers. The number of blue dots corresponds to the pentagonal number itself. Note the equal spacing when dealing with dots on each segment (eg: some interior blue dots are midpoints, others are quarter points, etc.)
<h3>Answer: Pentagonal number</h3>