Answer:
3 in wide by 4.5 in long.
Step-by-step explanation:
3/4=0.75
I divided the given dimensions by two, so I could multiply .75 by the sets of twos instead of dividing 3/4 by two and multiplying that by 12 and 8.
.74 x 4 = 3
.75 x 6 = 4.5
I believe in standard form this would be 31/2 + 27i/2
Answer:
19.44 hours, about 19 hours 26 minutes
Step-by-step explanation:
The exponential equation that describes your caffeine level can be written as ...
c(t) = 120·(1 -0.12)^t . . . . where t is in hours and c(t) is in mg
We want to find t for c(t) = 10, so ...
10 = 120(0.88^t)
10/120 = 0.88^t . . . . . . . divide by 120
log(1/12) = t·log(0.88) . . . take logarithms
t = log(1/12)/log(0.88) ≈ 19.4386
It will take about 19.44 hours, or 19 hours 26 minutes, for the caffeine level in your system to decrease to 10 mg.
Answer:
He has 3 pigs and 5 chickens
Step-by-step explanation:
Because pigs have 4 legs and 3 pigs would be 12 legs and then chickens have 2 legs so 5 chickens times 2 would be 10. Therefore he has 3 pigs and 5 chickens
:<span> </span><span>You need to know the derivative of the sqrt function. Remember that sqrt(x) = x^(1/2), and that (d x^a)/(dx) = a x^(a-1). So (d sqrt(x))/(dx) = (d x^(1/2))/(dx) = (1/2) x^((1/2)-1) = (1/2) x^(-1/2) = 1/(2 x^(1/2)) = 1/(2 sqrt(x)).
There is a subtle shift in meaning in the use of t. If you say "after t seconds", t is a dimensionless quantity, such as 169. Also in the formula V = 4 sqrt(t) cm3, t is apparently dimensionless. But if you say "t = 169 seconds", t has dimension time, measured in the unit of seconds, and also expressing speed of change of V as (dV)/(dt) presupposes that t has dimension time. But you can't mix formulas in which t is dimensionless with formulas in which t is dimensioned.
Below I treat t as being dimensionless. So where t is supposed to stand for time I write "t seconds" instead of just "t".
Then (dV)/(d(t seconds)) = (d 4 sqrt(t))/(dt) cm3/s = 4 (d sqrt(t))/(dt) cm3/s = 4 / (2 sqrt(t)) cm3/s = 2 / (sqrt(t)) cm3/s.
Plugging in t = 169 gives 2/13 cm3/s.</span>
