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AVprozaik [17]
1 year ago
14

Suppose a spherical balloon grows in such a way that after t seconds, its volume is V = 4 sqrt(t) cm3. What is the volume of the

balloon at time t = 169 seconds?: V =? . How fast is the volume changing after 169 seconds?
Mathematics
1 answer:
Arisa [49]1 year ago
4 0
:<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>
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A stop sign is shown. Each side measures 12.4 inches and the distance from any side to the opposite is 30 inches. Divide the sto
Daniel [21]

A stop sign has a total of 8 sides measuring 12.4 inches on each side and 30 inches for the distance of each sides.

 

Given the measurement, the rectangle's dimensions are as follows:

If divided horizontally: Length = 12.4 inches, width = 30 inches

If divided vertically: Length = 30 inches, width = 12.4 inches

 

From the divided rectangle, we can produce a 3-side equal trapezoid. In this case, we will have a uniform measurement of 12.4 inches on each side and 30 inches for the longer side.

3 0
2 years ago
Suri's age is 4 less than 3 times her cousin's age. Suri is 17 years old. Which method can be used to find c, her cousin's age?
Mila [183]

Answer:

7 years

Step-by-step explanation:

Suri's age=17 years

Her cousin's age =c

Suri's age is 4 less than 3 times her cousin's age

4 less than 3 times her cousin's age means subtract 4 from 3 times her cousin's age

17=3c - 4

Find c which is her cousin's age

Add 4 to both sides

17+4=3c - 4 + 4

17+4=3c

21=3c

Divide both sides by 3

21/3=3c/3

7=c

Therefore,

C= 7

Her cousin's age = 7 years

4 0
1 year ago
Ethan sells e candy bars for $2.50 apiece and Chloe sells c candy bars for $2.00 apiece to raise
natta225 [31]
2.50x = Ethan’s total sold -6 (15/2.50=6)
2x = Chloe’s total sold
2.50(12) = 30 - 6 = 24
2(12) = 24
They both sold 12 candy bars
8 0
2 years ago
Read 2 more answers
Ahmed has five more CDs than one-half the number of CDs Julia has. In this situation, what does c+ 5 represent?
KonstantinChe [14]
The c+5 should represent the five more cds that he has thats more than julias

6 0
1 year ago
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Which statement identifies how to show that j(x) = 11.6ex and k(x) = In (StartFraction x Over 11.6 EndFraction) are inverse func
Vadim26 [7]

Answer:

<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>

Step-by-step explanation:

Given the function  j(x) = 11.6e^x and k(x) = ln \dfrac{x}{11.6}, to show that both equality functions are true, all we need to show is that both  j(k(x)) and k(j(x)) equal x,

For j(k(x));

j(k(x)) = j[(ln x/11.6)]

j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}

j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)

j[(ln x/11.6)] = 11.6 * x/11.6

j[(ln x/11.6)] = x

Hence j[k(x)] = x

Similarly for k[j(x)];

k[j(x)] = k[11.6e^x]

k[11.6e^x] = ln (11.6e^x/11.6)

k[11.6e^x]  = ln(e^x)

exponential function will cancel out the natural logarithm leaving x

k[11.6e^x]  = x

Hence k[j(x)] = x

From the calculations above, it can be seen that j[k(x)] =  k[j(x)]  = x, this shows that the functions j(x) = 11.6e^x and k(x) = ln \dfrac{x}{11.6} are inverse functions.

4 0
1 year ago
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