Step-by-step explanation:
area of square = 1
area of a circle in square =
bu estimate ![p^ = X/n = 760/1000 = 0.76Therefore, [tex]\frac{\pi }{4} \approx 0.76](https://tex.z-dn.net/?f=p%5E%20%3D%20X%2Fn%20%3D%20760%2F1000%20%3D%200.76%3C%2Fp%3E%3Cp%3ETherefore%2C%20%3C%2Fp%3E%3Cp%3E%0A%5Btex%5D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%5Capprox%200.76)
b) if n tend to infinity
Answer:
At a combined speed of 6 in/min, it takes us 24 mins to clean the wall
Step-by-step explanation:
Since the question did not provide the speed with which each student cleans, we can make assumptions. This is so that we can solve the question before us
Assuming student 1 cleans at a speed of 2 inches per minute, student 2 cleans at a speed of 2½ inches per minute & student 3 cleans at a speed of 1½ inches per minute.
Let's list the parameters we have:
Height of wall (h) = 12 ft, Speed (student 1) = 2 in/min, Speed (student 2) = 2½ in/min, Speed (student 3) = 1½ in/min
Speed of cleaning wall = Height of wall ÷ Time to clean wall
Time to clean wall (t) = Height of wall ÷ Speed of cleaning wall
since students 1, 2 and 3 are working together, we will add their speed together; v = (2 + 2½ + 1½) = 6 in/min
1 ft = 12 in
Time (t) = h ÷ v = (12 * 12) ÷ 6 = 144 ÷ 6
Time (t) = 24 mins
Write a system of equations and solve.
x + y = 10
5x + 10y = 90
Multiply and combine equations
-5x - 5y = -50
5x + 10y = 90
------------------
5y = 40
y = 8
x = 2
There are 2 nickels and 8 dimes in the jar. Hope this helps! :)
The answer is 48 degrees because you add 42 and 90 and then subtract from 180 (total degrees of a triangle)
Given the figure of a regular pyramid
The base of the pyramid is a hexagon with a side length = 6
The lateral area is 6 times the area of one of the side triangles
So, the side triangle has a base = 6
The height will be:
![\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%5E2%3D6%5E2%2B%28%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%5Ccdot6%29%5E2%3D36%2B27%3D63%20%5C%5C%20h%3D%5Csqrt%5B%5D%7B63%7D%20%5Cend%7Bgathered%7D)
so, the lateral area =
![6\cdot\frac{1}{2}\cdot6\cdot\sqrt[]{63}=18\sqrt[]{63}](https://tex.z-dn.net/?f=6%5Ccdot%5Cfrac%7B1%7D%7B2%7D%5Ccdot6%5Ccdot%5Csqrt%5B%5D%7B63%7D%3D18%5Csqrt%5B%5D%7B63%7D)
