Answer:
(2k+1)(k+3)
Step-by-step explanation:
14k²+49k+21
simplify by dividing the whole equation by 7
2k²+7k+3
factorise
(2k+1)(k+3)
Answer:
$21.57
Step-by-step explanation:
4.5 (1.22) + 0.8 (1.70) + 4.6 (3.20) =
5.49 + 1.36 + 14.72
21.57
Answer: Yes, Brad has enough oranges.
Step-by-step explanation:
Given :
The orange required for one bowl = 12 oz
Thus orange required for 12 bowls = 
Oranges available with Brad for 12 bowls = 10 lb = 
(1lb=16oz)
As organges required is less than oranges available with Brad, thus Brad has enough oranges.
Answer:
$8.00
Step-by-step explanation:
The problem statement gives two relations between the prices of two kinds of tickets. These can be used to write a system of equations for the ticket prices.
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<h3>setup</h3>
Let 'a' and 'c' represent the prices of adult and children's tickets, respectively. The given relations can be expressed as ...
a - c = 1.50 . . . . . . . adult tickets are $1.50 more
175a +325c = 3512.5 . . . . . total revenue from ticket sales.
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<h3>solution</h3>
We are only interested in the price of an adult ticket, so we can eliminate c to give one equation we can solve for 'a'. Using the first equation, an expression for c is ...
c = a -1.50
Substituting that into the second equation, we have ...
175a +325(a -1.50) = 3512.50
500a -487.50 = 3512.50 . . . . . . simplify
500a = 4000 . . . . . . add 487.50
a = 8 . . . . . . . . . divide by 500
An adult ticket costs $8.
