Answer:
66 pounds of tomatoes
Step-by-step explanation:
A recipe code lasagna calls for 3 pounds of tomatoes to serve 5 people. A caterer wants to make enough lasagna to serve 110 people. How many pounds of tomatoes does he neeed
From the above question:
5 people = 3 pounds of tomatoes
110 people = x pounds of tomatoes
Cross Multiply
5 people × x pounds = 110 people × 3 pounds
x pounds = 110 people × 3 pounds/5 people
x pounds = 66 pounds of tomatoes
Answer:
satisfies given homogenous solution and the particular solution is
.
Step-by-step explanation:
Given homogeneous equation is,
where t>0 subject to ![y_1(t)=t^2, y_2(t)=\frac{1}{t}\hfill (1)](https://tex.z-dn.net/?f=y_1%28t%29%3Dt%5E2%2C%20y_2%28t%29%3D%5Cfrac%7B1%7D%7Bt%7D%5Chfill%20%281%29)
To verify,
- wheather
satisfies given homogenous equation, then both will satisfy
that is,
![t^2y_{1}^{''}-2y_1=t^2\times 2-2\times t^2=0](https://tex.z-dn.net/?f=t%5E2y_%7B1%7D%5E%7B%27%27%7D-2y_1%3Dt%5E2%5Ctimes%202-2%5Ctimes%20t%5E2%3D0)
![t^2y_{2}^{''}-2y_2=t^2\times 2t^{-3}-2\times t^{-1}=2t^{-1}-2t^{-1}=0](https://tex.z-dn.net/?f=t%5E2y_%7B2%7D%5E%7B%27%27%7D-2y_2%3Dt%5E2%5Ctimes%202t%5E%7B-3%7D-2%5Ctimes%20t%5E%7B-1%7D%3D2t%5E%7B-1%7D-2t%5E%7B-1%7D%3D0)
Thus
and
satisfies (1).
Now the wronskean,
![W(y_1, y_2)(x)=y_1y_{2}^{'}-y_{2}y_{1}^{'}=-t^2t^{-2}-2tt^{-1}=-3\neq 0](https://tex.z-dn.net/?f=W%28y_1%2C%20y_2%29%28x%29%3Dy_1y_%7B2%7D%5E%7B%27%7D-y_%7B2%7Dy_%7B1%7D%5E%7B%27%7D%3D-t%5E2t%5E%7B-2%7D-2tt%5E%7B-1%7D%3D-3%5Cneq%200)
Thus
and
are solution of (1) .
where ![D\equiv \frac{\partial }{\partial t}](https://tex.z-dn.net/?f=D%5Cequiv%20%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20t%7D)
![=-\frac{1}{2t^2}\frac{1}{1-\frac{D^2}{t^2}}()5t^2-1](https://tex.z-dn.net/?f=%3D-%5Cfrac%7B1%7D%7B2t%5E2%7D%5Cfrac%7B1%7D%7B1-%5Cfrac%7BD%5E2%7D%7Bt%5E2%7D%7D%28%295t%5E2-1)
![=-\frac{1}{2t^2}(1-\frac{D^2}{t^2})^{-1}(5t^2-1)](https://tex.z-dn.net/?f=%3D-%5Cfrac%7B1%7D%7B2t%5E2%7D%281-%5Cfrac%7BD%5E2%7D%7Bt%5E2%7D%29%5E%7B-1%7D%285t%5E2-1%29)
![=-\frac{1}{2t^2}(1-\frac{D^2}{t^2}+........)(5t^2-1)](https://tex.z-dn.net/?f=%3D-%5Cfrac%7B1%7D%7B2t%5E2%7D%281-%5Cfrac%7BD%5E2%7D%7Bt%5E2%7D%2B........%29%285t%5E2-1%29)
sincce ![D^2(5t^2-1)=10](https://tex.z-dn.net/?f=D%5E2%285t%5E2-1%29%3D10)
![=\frac{5}{t^4}+\frac{1}{2t^2}-\frac{5}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B5%7D%7Bt%5E4%7D%2B%5Cfrac%7B1%7D%7B2t%5E2%7D-%5Cfrac%7B5%7D%7B2%7D)
Hence the result.
Answer:
22/35
Step-by-step explanation:
width/length = 66/105 = 22/35
Answer:
Step-by-step explanation:
Alright, lets get started.
This is the combination of three figures semi circle, rectangle and triangle.
We will find the area of each figure and will add them.
The radius of the circle is half of the diameter, so radius is 6.
Area of semi circle will be : ![\frac{1}{2} \times 3.14 \times 6^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%203.14%20%5Ctimes%206%5E2)
Area of semi circle is : 56.52
Area of rectangle is: ![length \times width](https://tex.z-dn.net/?f=length%20%5Ctimes%20width)
So, area of rectangle is:![40\times 12=480](https://tex.z-dn.net/?f=40%5Ctimes%2012%3D480)
Area of triangle is : ![\frac{1}{2}\times base\times height](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20base%5Ctimes%20height)
So, area of triangle is :![\frac{1}{2} \times 12\times 5=30](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2012%5Ctimes%205%3D30)
Adding all these three areas, the area of above figure will be:
[tex]Area = 56.52+480+30
Hence Area is 566.52............... Answer
Answer:
A=a+b
2h
Step-by-step explanation: