Since 1 oz is equal to 28.4 g and 1 kg is equal to 1000 g that means that 3521.12 kg are equal to 100 oz
So first step is to simplify everything outside of the radicals.
6*2=12
:. The expression is
__ __
12*\| 8 * \| 2
Now we know that
__ __ __
\| 8 = \| 4 * \| 2
And
__ __
\| 2 * \| 2 = 2
And
__
\| 4 = 2
So if we incorporate what we know into the equation, we can figure it out.
So let's first expand the radical 8.
__ __ __
12*\| 4 * \| 2 * \| 2
Now by simplifying the radical four and combining the radical twos, we can get all whole numbers.
12*2*2
Which equals 48.
Answer:48
Let the fraction be 5/5
remaining= 50%*5/5-1/5= 2.5/5-1/5=1.5/5
percentage =1.5/5*100=30%
Answer:
50 pounds
Step-by-step explanation:
Dan and june mix two kind of feed for pedigreed dogs
Feed A worth is $0.26 per pound
Feed B worth is $0.40 per pound
Let x represent the cheaper amount of feed and y the costlier type of feed
x+y= 70..........equation 1
0.26x + 0.40y= 0.30×70
0.26x + 0.40y= 21.........equation 2
From equation 1
x + y= 70
x= 70-y
Substitutes 70-y for x in equation 2
0.26(70-y) + 0.40y= 21
18.2-0.26y+0.40y= 21
18.2+0.14y= 21
0.14y= 21-18.2
0.14y= 2.8
Divide both sides by the coefficient of y which is 0.14
0.14y/0.14= 2.8/0.14
y= 20
Substitute 20 for y in equation 1
x + y= 70
x + 20= 70
x= 70-20
x = 50
Hence Dan and june should use 50 pounds of the cheaper kind in the mix
Answer: The required solution is 
Step-by-step explanation:
We are given to solve the following differential equation :
where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.
From equation (i), we have
Integrating both sides, we get
![\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]](https://tex.z-dn.net/?f=%5Cint%5Cdfrac%7Bdy%7D%7By%7D%3D%5Cint%20kdt%5C%5C%5C%5C%5CRightarrow%20%5Clog%20y%3Dkt%2Bc~~~~~~%5B%5Ctextup%7Bc%20is%20a%20constant%20of%20integration%7D%5D%5C%5C%5C%5C%5CRightarrow%20y%3De%5E%7Bkt%2Bc%7D%5C%5C%5C%5C%5CRightarrow%20y%3Dae%5E%7Bkt%7D~~~~%5B%5Ctextup%7Bwhere%20%7Da%3De%5Ec%5Ctextup%7B%20is%20another%20constant%7D%5D)
Also, the conditions are
and
Thus, the required solution is
