Answer:
<em>Value</em><em> </em><em>of</em><em> </em><em>p</em><em> </em><em>is</em><em> </em><em>14</em>
Step-by-step explanation:
<em>HAVE A NICE DAY</em><em>!</em>
<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>
Answer:
what does 30 yldpri the linear relationship have do with it
Step-by-step explanation:
Answer:
ASA
Step-by-step explanation:
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Answer:
2. 1 8/9 gallons
Step-by-step explanation:
5 2/3 divided by 3
5.66666667/3 = 1.88888889
.88888889 = 8/9
1 + 8/9 = 1 8/9
:<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>
