Jada has $200 to spend on flowers for a school celebration. She decides that the only flowers that she wants to buy are roses and carnations. Roses cost $1.45 each and

carnations cost $0.65 each. Jada buys enough roses so that each of the 75 people

attending the event can take home at least one rose.

1. Write an inequality to represent the constraint that every person takes home at least

one rose.

2. Write an inequality to represent the cost constraint.

This can be determined by  using inequality concept. Hence inequality to represent the constraint that every person takes home at least  one rose will be, R ≥ 75 and for the cost constraint will be, 1.45R + 0.65C ≤ 200.

Given:

Jada has $200 to spend on flowers.Roses cost $1.45 and  carnations cost $0.65 each and 75 people  attending the event can take home at least one rose.

Jada has total amount  = $200

 Let the number of roses purchased by Jada = R roses

Total number of people who will attending  event  = 75 people

Therefore,cost of R roses at the rate of $1.45 each = 1.45R

Along with roses Jada purchased carnations = C

Now the cost of each carnations= $0.65

Therefore,cost of C carnations at the rate of $0.65 each = 0.65C

1). Inequality to represent the constraint that every person takes home at least  one rose will be,

     R ≥ 75

2). Inequality representing the cost constraint will be,

      1.45R + 0.65C ≤ 200

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link

1). R ≥ 75

2). 1.45R + 0.65C ≤ 200

Step-by-step explanation:

Total amount with Jada = $200

Let the number of roses purchased by her = R

Number of people = 75

Cost of R roses at the rate of $1.45 each = 1.45R

Along with roses she purchased carnations = C

Cost of carnations = $0.65 each

Total cost of Carnations = $0.65C

1). Inequality to represent the constraint that every person takes home at least  one rose,

R ≥ 75

2). Inequality representing the cost constraint will be,

    1.45R + 0.65C ≤ 200

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.