Answer:
10.5 %
<u>Skills needed: Financial Math Essentials</u>
Step-by-step explanation:
1) First, before getting started, let's assume the price of the product is 
. This variable will be used a lot throughout the problem ().
2) Marking a price above means increasing the price in order to make money off of the purchased product. When raising something by 
percent, the new price would be 
.
---> In this case, the price increased by 
percent.
This means that it would be: 
New price is: 
3) The shopkeeper is then offering a 
percent discount off of this marked price. When offering a 
percent discount price, the new price (with discount), expressed algebraically is: 
---> the expression above simplifies to 
In this case, 
, 
---> 
This means that , with discount, has been raised 
.
10.5 % is the profit percent
(The profit percent being the final marked up price - purchased price)
The standard form of a quadratic equation is ,
ax² + bx + c = 0.
And the formula to find the discriminant is b² - 4ac.
Here the first step is to change the given equation into standard form. So, add 1 to each sides of the equation. Therefore,
2x² – 9x + 2+1 = –1 + 1
2x² – 9x + 3 = 0
Next step is to compare the given equation with this equation to get the value of a, b and c.
After comparing the equations we will get a = 2, b = -9 and c = 3.
So, discriminant = b²- 4ac
=( -9)²-4 (2)(3)
= 81 - 24
= 57
So, discriminant of the given equation is 57.
57 is greater than 0 and square root of 57 will result real number.
So, the correct choice is C: The discriminant is greater than 0, so there are two real roots.
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
