Answer:
Step 1 Eliminate fractions by multiplying all terms by the least common denominator of all fractions.
Step 2 Simplify by combining like terms on each side of the inequality.
Step 3 Add or subtract quantities to obtain the unknown on one side and the numbers on the other.
Step 4 Divide each term of the inequality by the coefficient of the unknown. If the coefficient is positive, the inequality will remain the same. If the coefficient is negative, the inequality will be reversed.
Step 5 Check your answer.
Caution: you need to use the same units of measurement throughout. If the spring stretches by 21 cm when a 135 newton object is attached, then you must ask for the mass (in newtons) of a fish that would stretch the spring by 62.1 cm.
We will need to assume that the spring is not stretched at all if and when no object is attached to the spring.
Write the ratio
21.0 cm 135 newtons
------------- = --------------------
62.1 cm x
Solve this for x. This x value represents the mass of a fish that would stretch the spring by 62.1 cm. You can cancel "cm" in the equation above:
21.0 135 newtons
------ = --------------------
62.1 x
Then 21.0x = (62.1)(135 newtons). Divide both sides of this equation by 21.0 to solve it for x.
Answer:
Sample size minimum is 60
Step-by-step explanation:
given that you want to create a 99% confidence interval with a margin of error of .5.
The population standard deviation is equal to 1.5
i.e. 
Confidence level = 99%
Since population std deviation is known, we can use Z critical value for finding margin of error
Z critical value for 99% = 2.58
Margin of error = 
Equate this to 0.5 and solve for n
Answer:
Step-by-step explanation:
The complete question is
Water flows into a tank according to the rate F(t)= (t+6)/(1+t), and at the same time empties out at the rate E(t)= (ln(t+2))/(t+1), with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest galllon, is in the tank at time t=10 minutes.
Let C(t) be the amount of water in the tank at time t. We now that the rate of change of the tank is given by
![\frac{dC}{dt}=[\tex]rate at which water flows in- rate at which water flows out. Then [tex]\frac{dC}{dt}=\frac{t+6}{t+1}-\frac{\ln(t+2)}{(t+1)}[\tex]so, the desired expression is obtained by integrating with respect to t. This leads us to [tex]C(t) = \int \frac{t+1}{t+1}+ \frac{5}{t+1} - \frac{\ln(t+2)}{(t+1)} dt=t+ 5 \ln (|t+1|)-\int \frac{\ln(t+2)}{(t+1)} dt +C](https://tex.z-dn.net/?f=%5Cfrac%7BdC%7D%7Bdt%7D%3D%5B%5Ctex%5Drate%20at%20which%20water%20flows%20in-%20rate%20at%20which%20water%20flows%20out.%20%3C%2Fp%3E%3Cp%3EThen%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cfrac%7BdC%7D%7Bdt%7D%3D%5Cfrac%7Bt%2B6%7D%7Bt%2B1%7D-%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Eso%2C%20the%20desired%20expression%20is%20obtained%20by%20integrating%20with%20respect%20to%20t.%20%3C%2Fp%3E%3Cp%3EThis%20leads%20us%20to%20%3C%2Fp%3E%3Cp%3E%5Btex%5DC%28t%29%20%20%3D%20%5Cint%20%5Cfrac%7Bt%2B1%7D%7Bt%2B1%7D%2B%20%5Cfrac%7B5%7D%7Bt%2B1%7D%20-%20%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%20dt%3C%2Fp%3E%3Cp%3E%3Dt%2B%205%20%5Cln%20%28%7Ct%2B1%7C%29-%5Cint%20%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%20dt%20%2BC)
Unfortunately, the integral 
cannot be expressed using fundamental functions. So, the problem cannot have an specific function (if you are willing to know the complete answer, the integral of this function uses the polylogarithm function with n=2).
Since you want the exact amount of water at time, you need to give C a value, that is, you need to know an initial condition for the problem. This means, you need to know the amount of water in the tank at time 0
Answer:
B should be 3 inches long because the area of the square above is 48 inches squared, so that long side is probably 8. B is 3 inches long. As for A, this should be simple. since we know the longer side of the room has the area of 27, we can tell the length(longest side) is 9. 14 minus 9 is equal to 4. A's width is 4 inches long.
Step-by-step explanation:
for the rest, since we know one of the widths is 6 inches, we know that the length of the room with an area of 32 inches is 8. therefore, since 8 times 4 equals the number 32, C is 4 inches long. back to B. since we know that c is 4 inches long and that there is a length of 11 inches, we can tell the longest side of room A is 7. 4 Times 7 equals 28, so B is 28 inches squared.
