Answer:
Their weights are:
Dorris = 5 lbs
Borris = 7 lbs
Morris = 9 lbs
Step-by-step explanation:
Let the weights of the frogs be:
Dorris = d
Borris = b
Morris = m
given:
d + b = 12 - - - (1)
b + m = 16 - - - (2)
d + m = 14 - - - (3)
from (3)
d + m = 14
m = 14 - d - - - (4)
putting the value of m into equation 2:
b + m = 16
b + (14 - d) = 16
b - d + 14 = 16
b - d = 16 - 14
b - d = 2
b = 2 + d - - - - (5)
putting the value of 'b' in equation 5 into equation (1)
d + b = 12 - - - (1)
where: b = 2 + d - - - (5)
∴ d + (2 + d ) = 12
d + d + 2 = 12
2d = 12 - 2
2d = 10
∴ d = 10 ÷ 2 = 5
d = 5
putting the value of 'd' into eqn (5)
b = 2 + d - - - - (5)
where d = 5
b = 2 + 5 = 7
b = 7 lbs
to find 'm', let us use eqn (3)
d + m = 14 - - - (3)
where d = 5
∴ 5 + m = 14
m = 14 - 5 = 9
m = 9 lbs
Therefore their weights are:
Dorris = 5 lbs
Borris = 7 lbs
Morris = 9 lbs
Option 3.
1/2(16x6)=48
1/2(96)=48
1/2(96)=48
48=48
The factor is three. One also but that is true for any two numbers.
Answer:
12s.
Step-by-step explanation:
Given:
Hudson's velocity, Vh = 8.8 ft/s
Katie's velocity, Vk = 6.3 ft/s
Kate's head start distance, Dk = 30 ft
From the above, Katie and Hudson meet at th = tk = t
But, Dk = Dh + 30
Velocity = Distance/time
t × Vk = Dk
t × Vh = Dh
t × Vk + 30 = Dk
t × Vk + 30 = Vh × t
6.3t + 30 = 8.8t
t = 30/2.5
= 12 s
