Answer:
B. No. The product of two fractions is the product of the numerators divided by the product of the denominators.
Step-by-step explanation:
For A: A is clearly false because you can multiply any fraction and it doesnt matter what the denominator is.
For B: The first part is correct, which is that to multiply two fractions, they dont have to have like/common denominators, the explanation part is also correct. When we multiply we multiply the numerator to the numerator and the denominator to the denominator.
For C: C is false because the first part says yes and we know that you can multiply any 2 fractions regardless of denominators.
For D: For D the first part is correct however, the explanation section is false you dont multiply the numerator to the denominator.
For E: You don't have to find the equivalent fraction to multiply because you can do that afterward.
I hope this helps, have a blessed day! :D
Answer:
-0.9
Step-by-step explanation:
-0.9 × 100 = -90
Answer:
a. 1620-x^2
b. x=810
c. Maximum value revenue=$656,100
Step-by-step explanation:
(a) Total revenue from sale of x thousand candy bars
P(x)=162 - x/10
Price of a candy bar=p(x)/100 in dollars
1000 candy bars will be sold for
=1000×p(x)/100
=10*p(x)
x thousand candy bars will be
Revenue=price × quantity
=10p(x)*x
=10(162-x/10) * x
=10( 1620-x/10) * x
=1620-x * x
=1620x-x^2
R(x)=1620x-x^2
(b) Value of x that leads to maximum revenue
R(x)=1620x-x^2
R'(x)=1620-2x
If R'(x)=0
Then,
1620-2x=0
1620=2x
Divide both sides by 2
810=x
x=810
(C) find the maximum revenue
R(x)=1620x-x^2
R(810)=1620x-x^2
=1620(810)-810^2
=1,312,200-656,100
=$656,100
Answer:
Step-by-step explanation:
(A) The difference between an ordinary differential equation and an initial value problem is that an initial value problem is a differential equation which has condition(s) for optimization, such as a given value of the function at some point in the domain.
(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.
(C) Example of a second order linear ODE:
M(t)Y"(t) + N(t)Y'(t) + O(t)Y(t) = K(t)
The equation will be homogeneous if K(t)=0 and heterogeneous if 
Example of a second order nonlinear ODE:
(D) Example of a nonlinear fourth order ODE:
![K^4(x) - \beta f [x, k(x)] = 0](https://tex.z-dn.net/?f=K%5E4%28x%29%20-%20%5Cbeta%20f%20%5Bx%2C%20k%28x%29%5D%20%3D%200)
Answer:
x=20
Step-by-step explanation:
Square both sides
4+x/6=4
4+x=4*6
x=24-4
x=20
