I need help.

Question

The number of telephone connection C(n) that can be made among n telephone is given by the quadratic function C(n) =1/2(n^2-n). How many telephones are needed to make 66 connections?

A financial planner invested part of 50,000 at 7.5% simple annual and the rest at 6% simple annual interest. The total interest earned the first year was $3,240. Find the amount invested at each rate

Answer 1

1)

Number of connections n = 66.

Given quadratic function of the number of telephone C(n) =1/2(n²-n).

C(n) =(n/2)(n – 1)

Substitute n = 66

C(66) =(66/2)(66 - 1).

= 33(65)

= 2145

The number of telephones for 66 connections is 2145.

Answer 2

2)

Let

x = The number of dollars invested at simple annual interest rate of 7.5 %

y = The number of dollars invested at  simple annual interest rate of 6 %

Total investment = x + y

Actual Total investment = 50000.

x + y = 50000

y = 50000 - x.                              --------------------------------- (1)

Time = 1 year

Formula of simple interest I = PTR/100

Interest earned at the rate of 7.5 % is I_x = (x)(1)(7.5)/100

I_x = 0.075x

Interest earned at the rate of 6 % is I_y = (y)(1)(6)/100

I_y = 0.06y

The total interest earned the first year = I_x + I_y = 0.075x + 0.06y

Actual total interest earned the first year = 3240

0.075x + 0.06y = 3240

Multiply each side by 100.

( 0.075x + 0.06y ) 100 = (3240)(100)

7.5x + 6y = 324000.                           --------------------------------- (2)

Substitute equation (1) in equation (2)

7.5x + 6(50000 - x) = 324000

7.5x + 300000 - 6x = 324000

1.5x = 324000 – 300000

1.5x = 24000

x = 24000 / 1.5

x = 16000.

Substitute x = 16000 in equation (1)

y = 50000 - 16000

y = 34000.

Solution :

The $ 16,000 invested at simple annual interest rate of 7.5 %.

The $ 34,000 invested at simple annual interest rate of 6 %.