The answer is 130 degrees b
To find out what fraction of the half quart each friend will get, divide the 1/2 quart into 2 groups.
1/2 divided by 2 is the same as 1/2 x 1/2 or 1/4 quart.
Each friend will get 1/4 quart of milk.
Solve:
"<span>twice the number minus three times the reciprocal of the number is equal to 1."
3(1)
Let the number be n. Then 2n - ------- = 1
n
Mult all 3 terms by n to elim. the fractions:
2n^2 - 3 = n. Rearranging this, we get 2n^2 - n - 3 = 0.
We need to find the roots (zeros or solutions) of this quadratic equation.
Here a=2, b= -1 and c= -3. Let's find the discriminant b^2-4ac first:
disc. = (-1)^2 - 4(2)(-3) = 1 + 24 = 25.
That's good, because 25 is a perfect square.
-(-1) plus or minus 5 1 plus or minus 5
Then x = ------------------------------ = --------------------------
2(2) 4
x could be 6/4 = 3/2, or -5/4.
You must check both answers in the original equation. If the equation is true for one or the other or for both, then you have found one or more solutions.</span>
Answer:
197 in ^2 (answer B of the list)
Step-by-step explanation:
Notice that this figure has a total of 6 faces, four of which are rectangles (whose area is calculated as "base times height") and two trapezoids (whose area is (B+b)H/2 ).
The total surface area is therefore the addition of these six areas:
Rectangles:
5 in x 5 in = 25 in^2
5 in x 5 in = 25 in^2
5 in x 6.4 in = 32 in^2
9 in x 5 in = 45 in^2
Trapezoids:
Two of equal dimensions: B = 9 in, b = 5 in, H = 5 in
2 * (9 in + 5 in) 5 in /2 = 70 in^2
Which gives a total of (25 + 25 + 32+45 + 70) in^2 = 197 in^2
This agrees with answer B of he provided list.
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
