Only two problems I don’t understand. Don’t just give me answer, explain too please.
1 answer:
Answer:
12. Option b: 20 is the correct answer.
13. Option a: A rose costs $1 more than a daisy is the correct answer.
Step-by-step explanation:
12. Given that:
15x+10y=700 Eqn 1
x+y=60 Eqn 2
To find x, we will eliminate y be multiplying Eqn 2 by 10
10(x+y=60)
10x+10y=600 Eqn 3
Subtracting Eqn 3 from Eqn 2
(15x+10y)-(10x+10y)=700-600
15x+10y-10x-10y=100
5x=100
Dividing both sides by 5
Hence,
Option b: 20 is the correct answer.
13. Given that:
5x+4y=32 Eqn 1
x+6y=22 Eqn 2
Multiplying Eqn 2 by 5
5(x+6y=22)
5x+30y=110 Eqn 3
Subtracting Eqn 1 from Eqn 3
(5x+30y)-(5x+4y)=110-32
5x+30y-5x+4y=78
26y=78
Dividing both sides by 26
Putting y=3 in Eqn 2
x+6(3)=22
x+18=22
x=22-18
x=4
Hence,
Option a: A rose costs $1 more than a daisy is the correct answer.
You might be interested in
a. The first part asks for how many ways they can be seated together in a row. Therefore we want the permutations of the set of 6 people, or 6 factorial,
6! = 6 5
= 30 4
= 360 2 = 720 possible ways to order 6 people in a row
b. There are two cases to consider here. If the doctor were to sit in the left - most seat, or the right - most seat. In either case there would be 5 people remaining, and hence 5! possible ways to arrange themselves.
5! = 5 4
= 20 3
= 120 1 = 120 possible ways to arrange themselves if the doctor were to sit in either the left - most or right - most seat.
In either case there are 120 ways, so 120 + 120 = Total of 240 arrangements among the 6 people if the doctor sits in the aisle seat ( leftmost or rightmost seat )
c. With each husband on the left, there are 3 people left, all women, that we have to consider here.
3! = 3 2 6 ways to arrange 3 couples in a row, the husband always to the left