Given:
One midsegment of an equilateral triangle.
To find:
The ratio of the length of one midsegment of an equilateral triangle to the sum of two of its side lengths.
Solution:
All sides of an equilateral triangle are same.
Let a be the each side of the equilateral triangle.
Length of the midsegment is equal to the half of the non included side or third side.
![Midsegment=\dfrac{a}{2}](https://tex.z-dn.net/?f=Midsegment%3D%5Cdfrac%7Ba%7D%7B2%7D)
The sum of two side is
![a+a=2a](https://tex.z-dn.net/?f=a%2Ba%3D2a)
Now, the ratio of the length of one midsegment of an equilateral triangle to the sum of two of its side lengths is
![\text{Required ratio}=\dfrac{\text{Length of midsegment}}{\text{sum of two sides}}](https://tex.z-dn.net/?f=%5Ctext%7BRequired%20ratio%7D%3D%5Cdfrac%7B%5Ctext%7BLength%20of%20midsegment%7D%7D%7B%5Ctext%7Bsum%20of%20two%20sides%7D%7D)
![\text{Required ratio}=\dfrac{\dfrac{a}{2}}{2a}](https://tex.z-dn.net/?f=%5Ctext%7BRequired%20ratio%7D%3D%5Cdfrac%7B%5Cdfrac%7Ba%7D%7B2%7D%7D%7B2a%7D)
![\text{Required ratio}=\dfrac{a}{4a}](https://tex.z-dn.net/?f=%5Ctext%7BRequired%20ratio%7D%3D%5Cdfrac%7Ba%7D%7B4a%7D)
![\text{Required ratio}=\dfrac{1}{4}](https://tex.z-dn.net/?f=%5Ctext%7BRequired%20ratio%7D%3D%5Cdfrac%7B1%7D%7B4%7D)
![\text{Required ratio}=1:4](https://tex.z-dn.net/?f=%5Ctext%7BRequired%20ratio%7D%3D1%3A4)
Therefore, the ratio of the length of one midsegment of an equilateral triangle to the sum of two of its side lengths is 1:4.
Answer:
Domain = -2, range = -5
Domain =>0, range = 0
domain = 2, range = 5
let y =>f(x)
It can be seen that Function can be linear having common difference 2
F(x) = 2x + 1 for x>0
is satisfied for above values
let's check
F(2) = 2(2)+1 = 4+1 = 5 As given
For x<0 , F(x) = 2x -1
f(-2)= 2(-2) -1= -4-1= -5 As given
For X= 0, F(x)= 2x
So , F(0) = 0 As given
Answer: option D.
Step-by-step explanation:
The equation of the line in Slope-intercerpt form is:
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
Where "m" is the slope of the line and "b" is the intersection of the line with the y-xis.
The slopes of two perpendicular lines are negative reciprocals. Then, you need to find the slope of the line given in the graph, with the formula:
![m=\frac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
Then, this is:
![m=\frac{1-2}{0-(-3)}=-\frac{1}{3}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B1-2%7D%7B0-%28-3%29%7D%3D-%5Cfrac%7B1%7D%7B3%7D)
Then the slope of the other line is:
![m=3](https://tex.z-dn.net/?f=m%3D3)
Substittute the point (3,4) into
and solve for "b":
![4=3(3)+b\\4-9=b\\-5=b](https://tex.z-dn.net/?f=4%3D3%283%29%2Bb%5C%5C4-9%3Db%5C%5C-5%3Db)
Substituting you get that the equation of this line is:
![y=3x-5](https://tex.z-dn.net/?f=y%3D3x-5)
Answer:
The slope is
so the line would go up one over two instead of down one over two, therefore the graph would look like this:
If angle TSV=6x+17 then arc it intercept (SV)would be double it
6x+17=1/2 (15x+13)
12x+34=15x+13
21=3x
x=7
Arc SV=15(7)+13
=118
And angle TSV=6(7)+17
=59
There are 360 degrees in a circle and the measure of the arc would be equal to the central point so the sum of all the arcs equals 360
Arc SV+ arc SUV=360
118+SUV=360
Arc SUV=242