₦airaland Forum

Pridestorm:
Next Please

Mechanics96:
The sum to infinity of a geometric series is S. The sum to infinity of the squares of the terms is 2S. The sum to infinity of the cubes is 64/13 S. Find S and the first three terms of the original series

Pridestorm:
IS TOO ADVANCED FOR ME AT THIS MY LEVEL.......

Polymath56:
Original series: 8/3,8/9,8/27

S=4

Mechanics96:
While we await someone to come up with the solution to that,
Pridestorm, can you;
Convert 11.101 base two to base 10

Jackossky:
29/8 (improper fraction) 3.625 (decimal)

Jackossky:
29/8 (improper fraction) 3.625 (decimal)

mustapha112:
Next question

MasterMind21:
Then it should be 16, 65 and 321

Pridestorm:
x articles are produced at a total cost of $(2x"2 + 30x + 20), and each one is sold for $(x/3+100). Find the value of x which gives the greatest profit, and find this profit.

afo7219:
X=21 Profit =617

Pridestorm:
x articles are produced at a total cost of $(2x"2 + 30x + 20), and each one is sold for $(x/3+100). Find the value of x which gives the greatest profit, and find this profit.

Pridestorm:
x articles are produced at a total cost of $(2x"2 + 30x + 20), and each one is sold for $(x/3+100). Find the value of x which gives the greatest profit, and find this profit.

Polymath56:
x = 21 Profit = $715

Pridestorm:
Excellent!Excellent!

Pridestorm:
Okay.....

hostine316:
I want to know how the profit is $715

hostine316:
After solving for your X, you should put the value of X back in the proft equation (the equation that generated that value of X, i.e after checking for the critical point), which is -15x^2 +210x - 60.
I think I know where you got 715 from which i think is wrong. You dont insert X back into the cost and selling equations, rather it must be inserted into the profit equation to obtain the equivalent maximum profit.

If i am wrong, then you may have to thoroughly convince me because I haven't seen such approach you just used.

Polymath56:
Profit = x(x/3+100) - 2x²+30x+20
Profit = x²/3 + 100x - (2x²+30x+20)
Profit = -5x²/3 + 70x - 20
Profit is maximum when dp/dx = 0
dp/dx = -10x/3 + 70 = 0
x = 21

Profit = -5x²/3 + 70x - 20
Profit = -735+1470-20
Profit = $715

hostine316:
After solving for your X, you should put the value of X back in the proft equation (the equation that generated that value of X, i.e after checking for the critical point), which is -15x^2 +210x - 60.
I think I know where you got 715 from which i think is wrong. You dont insert X back into the cost and selling equations, rather it must be inserted into the profit equation to obtain the equivalent maximum profit.

If i am wrong, then you may have to thoroughly convince me because I haven't seen such approach you just used.

afo7219:
Next question,
Alhaji bello has 36 months car loan of #50,000 @ 12% interest rate per annum.what is the monthly loan repayment??

Pridestorm:
#18,000

Polymath56:
Total payable Interest for d 36mnts = N9786
Total cost of loan payment= N59786

Monthly Loan repayments = 59786/36 = N1661