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Zepler [3.9K]
2 years ago
11

If A and B are two angles in standard position in Quadrant I, find cos( A + B) for the given function values.

Mathematics
1 answer:
Naddik [55]2 years ago
3 0
You have to find the cos of angle A, so use the Pythagorean equation and trig laws to find the other side of the triangle created by angle A. 3^2 + x^2 = 5^2. x=4. This means cos(A) = 4/5). Make both cos (A) and cos (B) have equal denominators, and add. 148/185 + 60/185 = 208/185. This answer is correct, though it doesn’t appear to be any of the answers you wrote, so either those answers are wrong or you wrote something incorrectly in the problem.
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Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more than 120 in. Find
mafiozo [28]

Answer:

x = 20 in

L = 40 in

Step-by-step explanation:

Solution:-

- Denote the following:

 The side of square cross section = x

 The length of package = L

- Given that the combined length "L" of the package and girth "P" of the package must be less than and equal to 120 in

- The girth of the package denotes the Perimeter of cross section i.e square:

                       P = 4x

- The constraint for our problem in terms of combined length:

                      L + 4x ≤ 120  

                      L = 120 - 4x .... Eq1

- The volume - "V" -of the rectangular package with a square cross section is given as:

                      V = L*x^2   ... Eq2

- Substitute Eq1 into Eq2 and form a single variable function of volume "V":

                      V(x) = 120*x^2 - 4x^3

- We are asked to maximize the Volume - " V(x) " - i.e we are to evaluate the critical value of "x" by setting the first derivative of the Volume function to zero:

                     d [ V(x) ] / dx = 240x - 12x^2

                      240x - 12x^2 = 0

                      x*(240 - 12x) = 0

                      x = 0,  x = 20 in

- We will plug in each critical value of "x" back in function " V(x) ":

                      V (0) = 0

                      V(20) = 120(20)^2 - 4(20)^3

                                = 16,000 in^3

- The maximizing dimension of cross section is x = 20 in, the length of the parcel can be determined by the given constraint Eq1:

                       L = 120 - 4*20

                       L = 40 in

- The maximum volume of the rectangular package is with Length L = 40 in and cross section of Ax = ( 20 x 20 ):

5 0
3 years ago
What is Xcubed -(25cubed -2cubed)=. X=95
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Answer: I’m not 100% sure on the full answer but hopefully I can help you out a little. X cubed is x3

Step-by-step explanation:

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