Mean is were you add up all of the numbers and how many numbers are there thats how much you divide by and median is were you line up the numbers from least to greatest then try to find the middle number <span />
Answer:
D. Triangle QRS is an isosceles triangle because QR = RS.
Step-by-step explanation:
Find the length of each side of the triangle using the formula for calculating the distance between two points.
D = √(x2-x1)²+(y2-y1)²
For side RS
R(0,0) and S(5, -3.322)
RS = √(5-0)²+(-3.322-0)²
RS = √25+11.035684
RS = √36.035684
RS = 6.0029
For side RQ
R(0,0) and Q(-3, -5.2)
RQ = √(-3-0)²+(-5.2-0)²
RQ = √9+27.04
RQ = √36.04
RQ = 6.0033
For side QS
Q(-3,-5.2) and S(5, -3.322)
QS = √(5+3)²+(-3.322+5.2)²
QS = √64+3.526884
QS = √67.526884
QS = 8.22
From the calculation it can be seen that RS=QR
Since the two sides f the triangle are equal, hence the triangle is an isosceles triangle. An isosceles triangle is a triangle that has two of its sides equal
It is usual to represent ratios in their simplest form so that we are not operating with large numbers. Reducing ratios to their simplest form is directly linked to equivalent fractions.
For example: On a farm there are 4 Bulls and 200 Cows. Write this as a ratio in its simplest form.
Bulls <span>: </span>Cows
4 <span>: </span>200
If we halve the number of bulls then we must halve the number of cows so that the relationship between the bulls and cows stays constant. This gives us:
Bulls <span>: </span>Cows
2 <span>: </span>100
Halving again gives us
1 <span>: </span>50
So the ratio of Bulls to Cows equals 1 : 50. The ratio is now represented in its simplest form.
An example where we have 3 quantities.
On the farm there are 24 ducks, 36 geese and 48 hens.
Ratio of ducks <span>: </span>geese <span>: </span>hens
24 <span>: </span>36 <span>: </span>48
Dividing each quantity by 12 gives us
2 <span>: </span>3 : 4
So the ratio of ducks to geese to hens equals 2 : 3 : 4 which is the simplest form since we can find no further common factor.
first off let's notice that the height is 11 meters and the volume of the cone is 103.62 cubic centimeters, so let's first convert the height to the corresponding unit for the volume, well 1 meters is 100 cm, so 11 m is 1100 cm.
![\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=\stackrel{cm^3}{103.62}\\ h=\stackrel{cm}{1100} \end{cases}\implies 103.62=\cfrac{\pi r^2 (1100)}{3} \\\\\\ 3(103.62)=1100\pi r^2\implies \cfrac{3(103.62)}{1100\pi }=r^2 \\\\\\ \sqrt{\cfrac{3(103.62)}{1100\pi }}=r\implies \stackrel{cm}{0.00510199305952} \approx r](https://tex.z-dn.net/?f=%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20V%3D%5Cstackrel%7Bcm%5E3%7D%7B103.62%7D%5C%5C%20h%3D%5Cstackrel%7Bcm%7D%7B1100%7D%20%5Cend%7Bcases%7D%5Cimplies%20103.62%3D%5Ccfrac%7B%5Cpi%20r%5E2%20%281100%29%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%203%28103.62%29%3D1100%5Cpi%20r%5E2%5Cimplies%20%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%3Dr%5E2%20%5C%5C%5C%5C%5C%5C%20%5Csqrt%7B%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%7D%3Dr%5Cimplies%20%5Cstackrel%7Bcm%7D%7B0.00510199305952%7D%20%5Capprox%20r)