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Aleksandr-060686 [28]
3 years ago
13

What is 26364 of 58474337373745493

Mathematics
1 answer:
Delvig [45]3 years ago
6 0
Of means you're multipling. So just multiply them together :)
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A particular amuesment park ride will make 150 revolitions in 180 seconds at this rate how many revolutions will it make in 2 mi
Neko [114]

Answer:

100 revolts

Step-by-step explanation:

180s = 3min

150/3 is 50,

50 = 1 min

50 * 2 is 100

6 0
3 years ago
A marketing research firm would like to survey undergraduate and graduate college students about whether or not they take out st
neonofarm [45]

Answer:

Objective Minimize 10x1 +15x2 + 12x3+18x4+15x5+ 21x6

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

Step-by-step explanation:

Let

X1 = # of undergraduate students from the East region,

X2 = # of graduate students from the East region,

X3 = # of undergraduate students from the Central region,

X4 = # of graduate students from the Central region,

X5 = # of undergraduate students from the West region, and

X6 = # of graduate students from the West region.

Then  the cost functions are

y1= 10x1 +15x2

y2= 12x3+18x4

y3= 15x5+ 21x6

According to the given conditions

The constraints are

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500------- A

15X2 +18X4+21X6 ≥ 400---------B

21X6 ≥ 100

X6 ≥ 100/21

X6 ≥ 4.76

Taking

X6= 5

10X1 ≤ 500

X1 ≤ 500/10

X1≤ 5

18X4 ≥ 75

X4 ≥ 75/18

X4 ≥ 4.167

Taking

X4= 5

Putting the values

15X5+ 21X6 ≥ 300

15X5+ 21(5) ≥ 300

15X5+ 105 ≥ 300

15X5 ≥ 300-105

15X5 ≥ 195

X5 ≥ 195/15

X5 ≥ 13

Putting value of X6 and X4 in B

15X2 +18X4+21X6 ≥ 400

15X2 +18(5)+21(5) ≥ 400

15X2 +195 ≥ 400

15X2  ≥ 400-195

15X2  ≥ 205

X2  ≥ 205/15

X2  ≥ 13.67

Taking X2= 14

Now putting the values in the cost equations to check whether the conditions are satisfied.

y1= 10x1 +15x2

y1= 10 (5) + 15(14)= 50 + 210= $ 260

y3= 15x5+ 21x6

y3= 15 (13) + 21(5)

y3= 195+105= $ 300

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500

5+14+x3+5+13+5≥ 1500

x3≥ 1500-42

x3≥ 1458

y2= 12x3+18x4

y2= 12 (1458)  + 18 (5)

y2= 17496 +90

y2= $ 17586

The cost can be minimized if the number of students from

                                 Undergraduate         Graduate

East Region              X1≤ 5                            X2  ≥ 13.67

Central                      X3≥ 1458                   X4 ≥ 4.167

West                          X5 ≥ 13                          X6 ≥ 4.76

This will result in the required number of students that is 1500

Constraints:

East Undergraduate must not be greater than 5

East Graduate must not be less than 13

Central Undergraduate must  be greater than 1458

Central Graduate must  be greater than 4

West Undergraduate must  be greater than 13

West Graduate must  be greater than 4

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The east region has a least cost of $260 and west region has a cost of $300.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

6 0
3 years ago
N is prime given that it has 2 digits
rjkz [21]
Minimum 11, Maximum 97
8 0
3 years ago
Help me pls I am struggling In everything
kotegsom [21]

Answer:

The answers are A, B, C, F

3 0
3 years ago
Write the trigonometric expression sin(sin−1u−tan−1v) as an algebraic expression in u and v. Assume that the variables u and v r
igomit [66]

Answer:

[u – v√(1 – u²)]/√(1 + v²)

Step-by-step explanation:

Let sin^-1(u) = A, therefore sinA = u.

We know that sin(theta) = opposite/hypothenuse

Therefore, sinA = u/1 and u is the opposite side to angle A while 1 is the hypotenuse. Draw an acute triangle placing u opposite to angle A and 1 as the hypotenuse. By Pythagoras theorem the adjacent would be √(1 – u²).

By doing this, it means cosA = adjacent/hypotenuse = √(1 – u²)/1 = √(1 – u²)

Also, let tan^-1(v) = B, therefore tanB = v.

We know that tan(theta) = opposite/adjacent

Therefore, tanB = v/1 and v is the opposite side to angle B while 1 is the adjacent. Draw an acute triangle placing v opposite to angle B and 1 as the adjacent. By Pythagoras theorem the hypothenuse would be √(1 + v²).

Therefore, sinB = opposite/hypotenuse = v/√(1 + v²) and cosB = adjacent/hypotenuse = 1/√(1 + v²)

Now,

sin[sin^–1(u) – tan^–1(v)] =

sin(A – B) =

sinAcosB – sinBcosA =

u[1/√(1 + v²)] – [v/√(1 + v²)][√(1 – u²)] =

[u/√(1 + v²)] – [v√(1 – u²)/√1 + v²)] =

[u – v√(1 – u²)]/√(1 + v²).

8 0
3 years ago
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