Answer:
51 were the total number of the tickets sold at a school carnival were early-admission tickets.
Step-by-step explanation:
Total number of tickets sold by = 100
Let 
be the early-admission tickets sold by the school.
As 51% of the tickets sold at a school carnival were early-admission tickets.
so
Therefore, 51 were the total number of the tickets sold at a school carnival were early-admission tickets.
Kindly refer to attachment for solution.
Hope it helps ^_^
When a tangent line (13.5 cm) and a secant (lines x + 8.45 cm) intersect then:
tangent line^2 = 8.45 * (8.45 + x)
13.5^2 = 71.4025 + 8.45 x
182.25 -71.4025 = 8.45x
8.45 x = 110.8475
x = 13.1180473373
x = 13.1 (rounded)
Source:
1728.com/circangl.htm
I don’t know what your trying to find but you basically create an equation, like this
3x+5x=80
and solve which would result in x=10
so there is
30 geckos
and
50 anole lizards
Answer:
300 seconds
Step-by-step explanation:
The first dog run at v₁ = 15 m/sec the second one run at v₂ = 12 m/sec
we know that d = v*t then t = d/v
Then the first dog will take 300/ 12 = 25 seconds to make a turn
The second will take 300 / 15 = 20 seconds to make a turn
Then the first dog in 12 turns 12*25 will be at the start point, and so will the second one at the turn 15.
To check first dog 12 * 25 = 300
And the second dog 15 * 20 = 300
That means that time required for the two dogs to be at the start point together is
300 seconds, in that time the first dog finished the 12 turns, and the second had ended the 15.
Another procedure to solve this problem is as follows:
between 12 m/sec and 15 m/sec the minimum common multiple is 300 ( 300 is the smaller number that accept 12 and 15 as factors 12*15 = 300) Then when time arrives at 300 seconds the two dogs will be again in the starting point
