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Anna007 [38]
2 years ago
13

Find the range of the function for the domain(-4,-2,0, 1.5,4).f(x) = 5x²+ 4​

Computers and Technology
1 answer:
Masja [62]2 years ago
4 0
<h2>Hello!</h2>

The answer is:

The range or output for the given domain is:

(84,24,4,15.25,84)

<h2>Why?</h2>

To find the range (output) for the given domain (inputs) we need to evaluate the given function with the given inputs or "x" values.

So, evaluating, we have:

- Range with "x" equal to -4:

f(x)=5x^{2} +4\\\\f(-4)=5*(-4)^{2} +4=5*16+4=84

- Range with "x" equal to -2:

f(x)=5x^{2} +4\\\\f(-2)=5*(-2)^{2} +4=4*5+4=24

- Range with "x" equal to 0:

f(x)=5x^{2} +4\\\\f(-2)=5*(0)^{2} +4=0+4=4

- Range with "x" equal to 1.5:

f(x)=5x^{2} +4\\\\f(-2)=5*(1.5)^{2} +4=11.25+4=15.25

- Range with "x" equal to 4:

f(x)=5x^{2} +4\\\\f(-2)=5*(4)^{2} +4=80+4=84

Hence, the range or output for the given domain is:

(84,24,4,15.25,84)

Have a nice day!

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The mathematical constant Pi is an irrational number with value approximately 3.1415928... The precise value of this constant ca
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Answer:

I am writing a Python program:

def approxPIsquared(error):

   previous = 8

   new_sum =0

   num = 3

   while (True):

       new_sum = (previous + (8 / (num ** 2)))

       if (new_sum - previous <= error):

           return new_sum

       previous = new_sum

       num+=2    

print(approxPIsquared(0.0001))

Explanation:

I will explain the above function line by line.

def approxPIsquared(error):  

This is the function definition of approxPlsSquared() method that takes error as its parameter and approximates constant Pi to within error.

previous = 8     new_sum =0      num = 3

These are variables. According to this formula:

Pi^2 = 8+8/3^2+8/5^2+8/7^2+8/9^2+...

Value of previous is set to 8 as the first value in the above formula is 8. previous holds the value of the previous sum when the sum is taken term by term. Value of new_sum is initialized to 0 because this variable holds the new value of the sum term by term. num is set to 3 to set the number in the denominator. If you see the 2nd term in above formula 8/3^2, here num = 3. At every iteration this value is incremented by 2 to add 2 to the denominator number just as the above formula has 5, 7 and 9 in denominator.

while (True):  This while loop keeps repeating itself and calculates the sum of the series term by term, until the difference between the value of new_sum and the previous is less than error. (error value is specified as input).

new_sum = (previous + (8 / (num ** 2)))  This statement represents the above given formula. The result of the sum is stored in new_sum at every iteration. Here ** represents num to the power 2 or you can say square of value of num.

if (new_sum - previous <= error):  This if condition checks if the difference between the new and previous sum is less than error. If this condition evaluates to true then the value of new_sum is returned. Otherwise continue computing the, sum term by term.

return new_sum  returns the value of new_sum when above IF condition evaluates to true

previous = new_sum  This statement sets the computed value of new_sum to the previous.

For example if the value of error is 0.0001 and  previous= 8 and new_sum contains the sum of a new term i.e. the sum of 8+8/3^2 = 8.88888... Then IF condition checks if the

new_sum-previous <= error

8.888888 - 8 = 0.8888888

This statement does not evaluate to true because 0.8888888... is not less than or equal to 0.0001

So return new_sum statement will not execute.

previous = new_sum statement executes and now value of precious becomes 8.888888...

Next   num+=2  statement executes which adds 2 to the value of num. The value of num was 3 and now it becomes 3+2 = 5.

After this while loop execute again computing the sum of next term using       new_sum = (previous + (8 / (num ** 2)))  

new_sum = 8.888888.. + (8/(5**2)))

This process goes on until the difference between the new_sum and the previous is less than error.

screenshot of the program and its output is attached.

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