let us repersent the number of student tickets, then 's + 20' is repersenting the number of adult tickets.
The equation for the total tickets can be experssed as the following:
4.00 * s + 8.00 * ( s + 20 ) = 880.00
solving for 's=60', there were 60 student tickets.
60 + 20 = 80
there were 80 adult tickets.
Hope helps!-Aparri
Answer: The cost of a ticket bought online is $12.60.
Step-by-step explanation: First, we need to find the value of the transaction fee. Multiply 12 and 5%.
$12 x 0.05 = $0.60.
Now we have the fee. Lets add the fee with the original cost.
$12 + $0.60 = $12.60.
Therefore, we can conclude that the cost of a ticket that is purchased online is $12.60.
- What is the solution to the inequality -3x-42>3 ?
Add 42 to both sides.
Add 3 and 42 to get 45.
Divide both sides by -3. As -3 is <0, the inequality direction has changed.
Divide 45 by -3 to get -15.
Answer:
x = 5
Step-by-step explanation:
11 = 4x-9
20 = 4x
20/4 = 4x/4
5 = x
Answer:
Step-by-step explanation:
A1. C = 104°, b = 16, c = 25
Law of Sines: B = arcsin[b·sinC/c} ≅ 38.4°
A = 180-C-B = 37.6°
Law of Sines: a = c·sinA/sinC ≅ 15.7
A2. B = 56°, b = 17, c = 14
Law of Sines: C = arcsin[c·sinB/b] ≅43.1°
A = 180-B-C = 80.9°
Law of Sines: a = b·sinA/sinB ≅ 20.2
B1. B = 116°, a = 11, c = 15
Law of Cosines: b = √(a² + c² - 2ac·cosB) = 22.2
A = arccos{(b²+c²-a²)/(2bc) ≅26.5°
C = 180-A-B = 37.5°
B2. a=18, b=29, c=30
Law of Cosines: A = arccos{(b²+c²-a²)/(2bc) ≅ 35.5°
Law of Cosines: B = arccos[(a²+c²-b²)/(2ac) = 69.2°
C = 180-A-B = 75.3°
