The velocity of the particle is \[v(t)=s'(t)=3t^2-3t-2.\] We need to determine the value \(t\) for which \(v(t)=166\). This is equivalent to solving the equation \[3t^2-3t-168=0.\] We are allowed to use calculator, but we don't have to. The quadratic polynomial \(3t^2-3t-168\) can be factored as \[3t^2-3t-168=3\left(t^2-t-56\right)=3(t-8)(t+7).\] The solutions to the equation \(3(t-8)(t+7)=0\) are \(8\) and \(-7\). The only positive solution is \(8\). The correct answer is B.

The function \(y(x)\) has a horizontal tangent line at \(x\) if and only if \(y'(x)=0\). We need to calculate the derivative of \(y\). Then we need to solve the equation \(y'(x)=0\). We will first input the function into the memory. This is done by selecting \(\blacklozenge\) F1, and then we insert (3x^2+x-2)*e^(-x) under y1(x). Keep in mind that the exponential function \(e^x\) has to be accessed by typing \(\blacklozenge\) x. The derivative of y1(x) will be evaluated by the calculator and stored in y2(x). To achieve this, we go to the field y2(x) and first select the differential operator by typing 2ND 8. This will type the differential \(d\) and open bracket \((\). Inside the bracket we type y1(x),x). The obtained line becomes
y2(x)=\(d\)(y1(x),x)
After pressing the key enter the screen of the calculator will look like this: It is time to go back to the home screen by using the button HOME. Then we should type

solve(y2(x)=0,x)

and press \(\blacklozenge\) and then ENTER. The output screen looks like this: Therefore the solutions are \(x\approx -0.468375\) and \(x\approx 2.13504\). The correct answer is B.

Once we sell \(x\) items, we receive the revenue \(R(x)=6x\). The profit is \[P(x)=R(x)-C(x)=-0.01x^3+0.6x^2-7x.\] We will use the calculator TI-89 to find the maximum of the function \(P(x)\). We will first input the function into the memory. This is done by selecting \(\blacklozenge\) F1, and then we insert -0.01x^3+0.6x^2-7x under y1(x). (Notice that the first minus sign is obtained from the button (-) and the second minus sign is obtained by pressing the button -.) Then we press HOME button. On the home screen we give the command fMax(y1(x),x)|x>0. The command fMax is called by using F3 and then selecting fMax( from the menu. The sign > is obtained by choosing 2ND and then MATH. From the main menu, we select the option Test and then choose either of the symbols \(>\) or \(\geq\). After running the command fMax we obtain the following output: The function attains its maximum at \(x\approx 32.9099\). The maximum is equal to \(f(32.9099)\). On the home screen we can just print y1(32.9099) and obtain that the desired maximum is \(\approx 63.0331\). Therefore, the correct answer is A.

Let us denote by \(B\) the amount of the substance in the beginning. Then after time \(t\) the amount of substance is \[A(t)=B\cdot e^{\alpha t},\] for some constant \(\alpha\). We know that \(A(100)=\frac12B\). Therefore \[\frac{B}2=B\cdot e^{\alpha\cdot 100}.\] This implies that \(100\alpha=\ln\frac12\), which is equivalent to \(\alpha=\frac{\ln\frac12}{100}\). We need to determine \(t\) that solves the equation \[0.15B = B\cdot e^{\alpha t}.\] The last equation is equivalent to \[t=\frac{\ln(0.15)}{\alpha}=\frac{\ln(0.15)}{\frac{\ln\frac12}{100}}\approx 273.697.\] The correct answer is A.

The population after \(t\) years is given by the function \(P(t)\). The function \(P(t)\) has the exponential growth. It satisfies the equation \[P(t)=100000\cdot e^{0.045t}.\] We will now find the number \(t\) such that \(P(t)=200000\). This number \(t\) will represent the number of years that have to pass after January 1, 2016. The equation \[200000=100000\cdot e^{0.045t}\] is equivalent to \(2=e^{0.045t}\). We now take the logarthm of both sides and obtain \(\ln 2=0.045 t\). Therefore, \[t=\frac{\ln 2}{0.045}\approx 15.4033.\] \(15.4033\) years after January 1, 2016, the year will be \(2031\). The correct answer is B.

The amount of money after \(t\) years is a function \(M(t)\). The initial value \(M(0)\) satisfies \(M(0)=10000\). The function \(M(t)\) satisfies the equation \[M(t)=10000\cdot \left(1+\frac{0.158}{365}\right)^{365 t}.\] Thus, after \(2\) years, the amount of money is \[M(2)=10000\cdot \left(1+\frac{0.158}{365}\right)^{365 \cdot 2}\approx 13715.36.\] The rounding to the nearest dollar gives us \(13715\). The correct answer is C.

This problem is solved by using the calculator TI-89. We will use the exponential regression to determine the coefficients \(a\) and \(b\) such that the number of monthly text messages is approximated with the formula \[f(x)=a\cdot b^x.\] In the last formula \(x\) is the year and \(f(x)\) is the number of monthly text messages in the year \(x\).

• Step 1: Creation of the matrix. This is done with the following sequence of commands:

  APPS -> Data/Matrix Editor -> New

  The name of the matrix can be any word placed in the field Variable:. If you don't know what to choose, here is a suggestion: Give the name tm to your matrix. The name is short and easy to type. It also consists of the first letters of the name of the object "text message."
• Step 2: Fill the matrix with data. The following two pictures show you the obtained matrix. There are two pictures because the matrix has too many rows to fit on one screen. \(\quad\quad\)
• Step 3: Configure the regression. Use the function F5, then choose ExpReg option from the menu. Set the columns c1 and c2 for the coordinates x and y. In the line Store RegEq make a selection of y1(x). This will store the obtained function in the memory location y1(x).
• Step 4: Run the regression. You just need to press enter on the calculator and wait to see the coefficients \(a\) and \(b\). As you can see the coefficients are \(a=2.411344E-666\) and \(b=2.134173\). The function is \(f(x)=a\cdot b^x\).
• Step 5: Evaluate the function for the desired input parameter. Go to the home screen and request the calculator to evaluate y1(2017). The final result is 133.212. Therefore, the correct answer is D.

The variable \(x\) represents the number of products that are being sold. The variable \(p\) represents the price. The market equilibrium is a pair \((x_e,p_e)\) that solves the system of equations \begin{eqnarray*} p&=&35-x^2\\ p&=&3+x^2. \end{eqnarray*} Adding the two equations gives us \(2p=38\), hence \(p=19\). Now from the first equation we get \(x^2=16\). Since only positive solution can correspond to a number of products, then we get \(x=4\). Thus, the market equilibrium is \[(x_e,p_e)=(4,19).\] The producer surplus is given by the formula \[\Sigma_P=x_ep_e-\int_0^{x_e}S(x)\,dx,\] where \(S(x)=3+x^2\) is the supply function. The value of the integral is \begin{eqnarray*} \Sigma_P&=&4\cdot 19-\int_0^4\left(3+x^2\right)\,dx=4\cdot 19-\left(3x+\frac{x^3}3\right)_{x=0}^{x=4} \\&=&4\cdot 19-3\cdot 4-\frac{4^3}3\\ &\approx&42.667. \end{eqnarray*} Alternatively, we can use the calculator to evaluate the expression \(4\cdot 19-\int_0^4\left(3+x^2\right)\). We need to type 4*19-, then F3, then select the option 2: \(\int\) integrate, and then type 3+x^2,x,0,4).

The correct answer is B.

We can minimize the function by typing F3 and then selecting fMin(. Then we need to write the expression 2x-3+5/x,x)|x > 0. The symbol > is obtained by pressing the button 2ND and then the button 5 to access the MATH menu. From the math menu we need to choose the submenu 8. Test which contains the comparison operator >. The evaluation of the function gives us that the value of \(x\) is \(1.58\). The correct answer is E.

Let us define the functions \(y_1(x)=x^3-8x^2+18x-5\) and \(y_2(x)=x+5\). The area \(A\) is equal to the sum of the integrals \begin{eqnarray*} A&=&\int_1^2\left(y_1(x)-y_2(x)\right)\,dx+\int_2^5\left(y_2(x)-y_1(x)\right)\,dx. \end{eqnarray*} We can evaluate these integrals using the calculator. This is an example of a problem in which we will save a lot of time by placing the functions \(y_1(x)\) and \(y_2(x)\) int he memory. The functions are placed in the memory by typing \(\blacklozenge\) F1. In the slot y1(x) we insert x^3-8x^2+18x-5 and in the slot y2(x) we type x+5. Then we press the button HOME. We will evaluate the sum of the integrals by first typing F3 and then choosing the integral from the menu. The characters \(\int(\) will appear in the input line. We complete the input line by writing y1(x)-y2(x),x,1,2). Then we press the button +. The next step is to write another integral (F3 and then choose integral from the menu). The input line will receive the characters \(\int(\). We need to add y2(x)-y1(x),x,2,5). Since we need the approximate evaluation, we press the button \(\blacklozenge\) before the button ENTER. The value of the integral is \(11.833\). Therefore, the correct answer is B.