A good option available to you as a financial expert is to request a part payment from the customer while the remaining part is to be made on the next pay day.
For a good financial management, the best decision to make profit or avoid loss is paramount.
The most difficult challenge faced today by loan companies is the non complaisance of the customers to pay back when the period given expires.
The best option that may be available for the loan manager must be a request for the customer to pay part of the loan and complete the payment in the next pay day.
Thus, a good option available to you as a financial expert is to request a part payment from the customer while the remaining part is to be made on the next pay day.
Learn more about options for loan payment here: brainly.com/question/12233464
Step-by-step explanation:
sin ∅ + 1 = cos2∅
From multiple angles in trigonometry
cos2∅ = 1 - 2sin²∅
So we have
sin∅ + 1 = 1 - 2sin²∅
sin∅ +1 - 1 + 2sin²∅= 0
sin∅ + 2sin²∅ = 0
Factor out sin∅
sin∅ ( 1 + 2sin∅) = 0
sin ∅ = 0
∅ = 0° , 180°
1 + 2sin∅ = 0
2sin∅ = - 1
sin ∅ = -1/2
∅ = 210° , 330°
The solutions are
<h3>0° , 180° , 210° , 330°</h3>
Hope this helps you
Let's start by solving this problem when there are only two positive numbers involved, and then see whether we can apply the same technique when there are three positive integers.
Let the two positive integers be x and y.
Then x + y = 100, and xy = the product.
Let's eliminate x. Solve x + y = 100 for x: x = 100 - y. Now subst. this last result into P = xy: P = product = (100 - y)(y) = 100y - y^2
Differentiating, dP/dy = 100 - 2y. Set this = to 0 and solve for y: -2y = -100, and y = 50. Since x + y = 100, x is thus also = to 50.
Solution set: (50,50).
Now suppose that three positive integers add up to 100, and that we want to maximize their product.
Then x + y + z = 100. Let's maximize f(x,y,z) = xyz (the product of x, y and z).
Since x + y + z = 100, we can eliminate z by solving x + y + z = 100 for z and subst. the result back into f(x,y,z) = xyz:
We get f(x,y) =xy(100-x-y), a function of two variables instead of three.
I won't go through the entire procedure of maximizing a function in three variables, but will get you started:
Find the 'partial of f with respect to x' and then the 'partial of f with respect to y'. Set each of these partial derivatives = to 0:
f = 0 = (partial of xy(100-x-y) with respect to x
x
= xy(partial of 100-x-y with respect to x) + (100-x-y)(partial of xy with respect to x)
= xy(-1) + (100-x-y)(y)
We must set this partial = to 0: -xy+100y-xy-y^2 = 0
-2xy + 100y - y^2 = 0
or y(-2x + 100 - y) = 0
of which y=0 is one solution and in which -2x + 100 - y = 0
You must now go through the same procedure with respect to the partials with respect to y.
If you'd like to continue this discussion, please respond with questions and comments.
STEP 1
Find miles per minute; divide total miles by number of minutes
= 1.5 miles ÷ 35 minutes
= 0.0428571429 miles per minute
STEP 2
find miles per hour; multiply step 1 answer by 60 minutes per hour
= 0.0428571429 * 60
= 2.5714285714 miles per hour
ANSWER: Fifi walked 2.57 miles per hour
Hope this helps! :)
Answer:
+
Step-by-step explanation: