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Anarel [89]
3 years ago
15

An object dropped from a height of 600 feet has a height, h(t), in feet after t seconds have elapsed, such that h(t)=600 - 16t^2

. Express t as a function of height h, and find the time to reach a heigh of 50 feet
Mathematics
1 answer:
gulaghasi [49]3 years ago
7 0

Answer:

t as a function of height h is  t = √600 - h/16

The time to reach a height of 50 feet is 5.86 minutes

Step-by-step explanation:

Function for height is h(t) = 600 - 16t²

where t = time lapsed in seconds after an object is dropped from height of 600 feet

t  as a function of height h

replacing the function with variable h

h = 600 - 16t²

Solving for t

Subtracting 600 from both side

h - 600 = -16t²

Divide through by -16

600 - h/ 16 = t²

Take square root of both sides

√600 - h/16 = t

Therefore, t = √600 - h/16

Time to reach height 50 feet

t = √600 - h/16

substituting h = 50 in the equation

t = √600 - 50/16

t = √550/16

t= 34.375

t = 5.86 minutes

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Answer:

Part A. {S} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Part B. {A} = {1324, 1342, 1423. 1432}

Part C. {B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

Part D.

(A∪B) = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∩B) = ∅

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Step-by-step explanation:

A. All possible outcomes

There are four teams, each play a semi final where 1 and 2 plays against each other while 3 and 4 plays against each other. Winner of the first semi final can be either 1 or 2 therefore they both can not be in the championship game or in the losers game at the same time same goes for the other semi final.

Using this explanation (1324 denotes: 1 beats 2 and 3 beats 4 in first-round games and then 1 beats 3 and 2 beats 4), All possible outcomes are

{S} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

B. Event A in which 1 wins the tournament

From {S} we only have to write the outcomes in which 1 is the first number in 4digit combinations given in part A

{A} = {1324, 1342, 1423. 1432}

C. Event B in which 2 gets into championship game

From {S} we only have to write the outcomes in which 2 is either the first or second digit in 4digit combinations given in part A

{B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

D. Outcomes in (A∪B), (A∩B) and A'

I. (A∪B)

(A∪B) means A union B therefore all we have to do is combine all the members of A and B

(A∪B) = {1324, 1342, 1423, 1432} ∪{2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∪B) = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

II. (A∩B)

(A∩B) means A intersection B in which we have to find the common members of A and B. If there are no common members then the result of (A∩B) is a null set.

(A∩B) =  {1324, 1342, 1423, 1432} ∩ {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∩B) = ∅

III. {A'}

A' means A compliment, in other words it can be described as all the possible outcomes that are not part of A. So all we do is to subtract outcomes of A from the total possible outcomes S

{A'} =  {S} - {A}

{A'} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231} - {1324, 1342, 1423. 1432}

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

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