The number of cans needed to paint an area covering 2,200 sq. units will be given by:
number of cans=[area required to be painted]/[area that 1 can of paint can cover]
=2200/400
=5.5
Therefore we conclude that we will need 5.5 cans of paint
Answer:
+1/7
Step-by-step explanation:
One approach to evaluating this would be to rewrite it as:
-2 -5
------ * --------- .
10 7 This reduces to:
2
--------- = +1/7
2(7)
Answer:
the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively
Step-by-step explanation:
the equation that governs the remaining mass m of unobtanium -41 after a time t is
m=m₀*2^(-t/T) , where t is in seconds ,m₀ represents initial mass and T=half-life
Therefore
a) for t=5 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-5 s/5 s) = 48 gr/2 = 24 gr
b) for t=10 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-10 s/5 s) = 48 gr/4 = 12 gr
b) for t=15 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-15 s/5 s) = 48 gr/9 = 5.33 gr
b) for t=20 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-20 s/5 s) = 48 gr/16 = 3 gr
thus the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively
Answer:
7.6 seconds
Step-by-step explanation:
To solve this problem we can use the following equation:
S = So + 113*t + at^2/2
Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.
In this problem, we have that S = 0, So = 65, Vo = 113 and a = -32 ft/s2 (The acceleration of gravity)
So we have that:
0 = 65 + 113t - 16t2
16t2 - 113t - 65 = 0
Using Bhaskara's formula, we have:
Delta = 113^2 + 4*65*16 = 16929
sqrt(Delta) = 130.11
t = (113 + 130.11) / (2*16) = 7.597 seconds
Rounding to nearest tenth, we have t = 7.6 seconds
