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ZanzabumX [31]
3 years ago
10

3n^2-1 find the 2nd and 5th term

Mathematics
2 answers:
Butoxors [25]3 years ago
7 0

Answer:

2nd term = 11

5th term = 74

Step-by-step explanation:

3n² - 1

n = 2 , 2nd term = 3*2²  - 1

                           = 3*4 -1

                          = 12 - 1

                          = 11

n= 5 , 5th term = 3*5² - 1

                        = 3*25 - 1

                      = 75 - 1

                      = 74

user100 [1]3 years ago
5 0

Step-by-step explanation:

in\tt{3n^2-1  } ⠀

If n=2

so,

2nd term=⠀

\rightsquigarrow \tt{3×2^2-1  } ⠀

\rightsquigarrow \tt{ 3×4-1 } ⠀

\rightsquigarrow \tt{ 12-1 } ⠀

\rightsquigarrow \tt{ 11 } ⠀

<h3>so,2nd term is 11.</h3><h3 />

If n=5

so,

5th term =

\rightsquigarrow \tt{ 3×5^2-1 } ⠀

\rightsquigarrow \tt{ 3×25-1 } ⠀

\rightsquigarrow \tt{75-1  } ⠀

\rightsquigarrow \tt{74  }

<h3>so 5th term is 74.⠀</h3>
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Could you help me to solve the problem below the cost for producing x items is 50x+300 and the revenue for selling x items is 90
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Answer:

Profit function: P(x) = -0.5x^2 + 40x - 300

profit of $50: x = 10 and x = 70

NOT possible to make a profit of $2,500, because maximum profit is $500

Step-by-step explanation:

(Assuming the correct revenue function is 90x−0.5x^2)

The cost function is given by:

C(x) = 50x + 300

And the revenue function is given by:

R(x) = 90x - 0.5x^2

The profit function is given by the revenue minus the cost, so we have:

P(x) = R(x) - C(x)

P(x) = 90x - 0.5x^2 - 50x - 300

P(x) = -0.5x^2 + 40x - 300

To find the points where the profit is $50, we use P(x) = 50 and then find the values of x:

50 = -0.5x^2 + 40x - 300

-0.5x^2 + 40x - 350 = 0

x^2 - 80x + 700 = 0

Using Bhaskara's formula, we have:

\Delta = b^2 - 4ac = (-80)^2 - 4*700 = 3600

x_1 = (-b + \sqrt{\Delta})/2a = (80 + 60)/2 = 70

x_2 = (-b - \sqrt{\Delta})/2a = (80 - 60)/2 = 10

So the values of x that give a profit of $50 are x = 10 and x = 70

To find if it's possible to make a profit of $2,500, we need to find the maximum profit, that is, the maximum of the function P(x).

The maximum value of P(x) is in the vertex. The x-coordinate of the vertex is given by:

x_v = -b/2a = 80/2 = 40

Using this value of x, we can find the maximum profit:

P(40) = -0.5(40)^2 + 40*40 - 300 = $500

The maximum profit is $500, so it is NOT possible to make a profit of $2,500.

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