Answer:
24 cm²
Step-by-step explanation:
You can show your sister how to use the formula for the area of a triangle.
The base dimension is substituted for the letter b, the height is substituted for the letter h, and then the arithmetic is carried out.
A = 1/2bh
A = (1/2)(12)(4) = (6)(4) = 24 . . . . square centimeters
Using the combination formula, it is found that there are 495 ways to choose a 4-topping sandwich.
The order in which the toppings are chosen is not important, hence the <em>combination formula</em> is used to solve this question.
<h3>What is the combination formula?</h3>
is the number of different combinations of x objects from a set of n elements, given by:
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
4 toppings are chosen from a set of 12, hence the number of ways is given by:
![C_{12,4} = \frac{12!}{4!8!} = 495](https://tex.z-dn.net/?f=C_%7B12%2C4%7D%20%3D%20%5Cfrac%7B12%21%7D%7B4%218%21%7D%20%3D%20495)
More can be learned about the combination formula at brainly.com/question/25821700
#SPJ1
I'm not sure what the question is but I think your asking 150-85=65
Answer:
Therefore the diameter of the hole is
m.
Step-by-step explanation:
Bernoulli's equation,
![P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2](https://tex.z-dn.net/?f=P_1%2B%5Cfrac12%20%5Crho%20v%5E2_1%2B%5Crho%20g%20h_1%3D%20P_2%2B%5Cfrac12%20%5Crho%20v%5E2_2%2B%5Crho%20g%20h_2)
P₁ = P₂= atmospheric presser
= density
[since P₁ = P₂]
![\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)](https://tex.z-dn.net/?f=%5CRightarrow%5Crho%20%28%5Cfrac12%20v%5E2_1%2B%20g%20h_1%29%3D%20%5Crho%28%5Cfrac12%20v%5E2_2%2B%20g%20h_2%29)
![\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2](https://tex.z-dn.net/?f=%5CRightarrow%5Cfrac12%20v%5E2_1%2B%20g%20h_1%3D%20%5Cfrac12%20v%5E2_2%2B%20g%20h_2)
![\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2](https://tex.z-dn.net/?f=%5CRightarrow%5Cfrac12%20v%5E2_2-%5Cfrac12%20v%5E2_1%3Dg%20h_1-%20g%20h_2)
[
]
Here ![v_1\approx 0](https://tex.z-dn.net/?f=v_1%5Capprox%200)
![\Rightarrow v^2_2=2g h](https://tex.z-dn.net/?f=%5CRightarrow%20v%5E2_2%3D2g%20h)
![\therefore v_2=\sqrt {2gh](https://tex.z-dn.net/?f=%5Ctherefore%20v_2%3D%5Csqrt%20%7B2gh)
Here g= 9.8 m/s² , h = 10.4 m
The velocity of water that leaves from the hole
=
m/s
=14.28 m/s.
Given, the rate of flow from the leak is
![m^3/s](https://tex.z-dn.net/?f=m%5E3%2Fs)
Let the diameter of the hole be d.
Then the cross section area of the hole is ![=\pi (\frac d2)^2](https://tex.z-dn.net/?f=%3D%5Cpi%20%28%5Cfrac%20d2%29%5E2)
We know that,
The rate of flow = Cross section area × speed
![\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B2.53%5Ctimes%2010%5E%7B-3%7D%7D%7B60%7D%20%3D%5Cpi%20%28%5Cfrac%20d2%29%5E2%5Ctimes%2014.28)
![\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}](https://tex.z-dn.net/?f=%5CRightarrow%20%28%5Cfrac%20d2%29%5E2%3D%5Cfrac%7B2.53%5Ctimes%2010%5E%7B-3%7D%7D%7B60%5Ctimes%2014.28%5Ctimes%20%5Cpi%7D)
![\Rightarrow d= 1.94 \times 10^{-3}](https://tex.z-dn.net/?f=%5CRightarrow%20d%3D%201.94%20%5Ctimes%2010%5E%7B-3%7D)
Therefore the diameter of the hole is
m.