A company has $8,970 available per month for advertising. Newspaper ads cost $200 each and can't run more than 19 times per month. Radio ads cost $470 each and can't run more than 28 times per

month at this price.

Each newspaper ad reaches 5800 potential customers, and each radio ad reaches 7100 potential

customers. The company wants to maximize the number of ad exposures to potential customers.

Use n for number of Newspaper advertisements and r for number of Radio advertisements.

Maximize P =

subject to

< 19

< 28

< $8,970

Enter the solution below. If needed, round ads to 1 decimal palace and group exposure to the

nearest whole person.

Number of Newspaper ads to run is

Number of Radio ads to run is

Maximum target group exposure is

people

go on the tube you might find some information there

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w

$655.61 which is option c

step-by-step explanation:

price of house =   $146,000

initial rate = 3.5%

mortgaged paid in this = 3.5 % of $146,000

                                        =

                                        =$5110

the total years would be = 7

total mortgaged paid = 7 * $5110

                                      =$35770

remaining money = $146000-35770

                                =$110230

new mortgaged payment = 5 % of 110230

                                            =  

                                            =$655.61