It would have taken you 5:30 for each mile. 55/10 is 5.5 thats 5 and a half, half a minute is 30 seconds. so 5 minutes and 30 seconds
We can solve this problem by stating that the summation
of momentums must be zero. That is, the momentum at one end of the pivot must
be equal to the momentum at the other end.
Momentum 1 = Momentum 2
Since Momentum is the product of Force and Distance, and
with this rule created, we can say that:
F1 * d1 = F2 * d2
Where,
F1 = 60 lbs
d1 = 2 inches from the pivot
F2 = unknown X
d2 = 3 inches from the pivot
Substituting to the equation to find for F2:
60 lbs * 2 inches = F2 * 3 inches
F2 = 40 lbs
<span>Therefore 40 lbs of upward force must be pushed to open
the valve.</span>
<span>Answer: B</span>
Answer:
The required division problem he must solve is:

Step-by-step explanation:
Consider the provided information.
Martin chose two of the cards below. When he found the quotient of the numbers, his answer was -16/9.
As we know that the quotient of the number is a negative number.
Therefore, the sign of both numbers must be different,
Thus we can concluded he must select
as one of the card, so that product is a negative number.
Let the selected card be x.

Hence, the two cards should be
and 
The required division problem he must solve is:

Let F be the father's current age, and E be Evie's current age.
F = E + 36 this is the current relation between the ages
F + 3 = 5(E + 3) this the relationship in 3 year (hence the + 3)
Then solve by substituting one equation into another:
(E + 36) + 3 = 5(E + 3)
E + 39 = 5E + 15
24 = 4E
6 = E
Evie's current age is 6.
Let m∠CLN = x. Then m∠ALM = 3x, and m∠A = 90°-x, m∠C = 90°-3x.
The sum of angles of ∆ABC is 180°, so we have
... 180° = 40° + m∠A + m∠C
Using the above expressions for m∠A and m∠C, we can write ...
... 180° = 40° + (90° -x) + (90° -3x)
... 4x = 40° . . . . . . . . . add 4x-180°
... x = 10°
From which we conclude ...
... m∠C = 90°-3x = 90° - 3·10° = 60°
The ratio of CN to CL is
... CN/CL = cos(∠C) = cos(60°)
... CN/CL = 1/2
so ...
... CN = (1/2)CL