**[Calc]**** ****Question*** *** Easy**

\(1210(x+120)=120\)

Which of the following equations has the same solution as the given equation?

A. \(x+120=12\)

B. \(x+120=130\)

c. \(x+12=12\)

D. \(x+12=120\)

**▶️Answer/Explanation**

Ans:A

To find an equation with the same solution as \(10(x+120) = 120\), we first need to solve the given equation for \(x\).

\[10(x+120) = 120\]

Divide both sides by \(10\):

\[x + 120 = 12\]

Subtract \(120\) from both sides:

\[x = 12 – 120\]

\[x = -108\]

Now, let’s check which of the given equations has the same solution:

A. \(x + 120 = 12\) -> \(x = -108\)

B. \(x + 120 = 130\) -> \(x = 10\)

C. \(x + 12 = 12\) -> \(x = 0\)

D. \(x + 12 = 120\) -> \(x = 108\)

Among these options, option A has the same solution as the given equation, \(x = -108\).

Therefore, the correct answer is option A: \(x + 120 = 12\).

*Question*

If \(x\)>0, which of the following is equivalent to \(\frac{1}{x}+\frac{1}{2x}\)

- \(\frac{1}{x}\)
- \(\frac{1}{2x}\)
- \(\frac{3}{2x}\)
- \(\frac{2}{3x}\)

**Answer/Explanation**

Ans: C

*Question*

There were no jackrabbits in Australia before 1788 when 24 jackrabbits were introduced, By 1920 the population of jackrabbits had reached 10 billion. If the population had grown exponentially, this would correspond to a 16.2% increase, on average, in the population each year. Which of the following functions best models the population \(p(t)\) of jackrabbits \(t\) years after 1788?

- \(p(t)=1.162(24)^t\)
- \(p(t) = 24(2 )^{1.162t}\)
- \(p(t)=24(1.162)^t\)
- \(p(t) = (24. 1.162 )^t\)

**Answer/Explanation**

Ans: C

*Question*

An initial investment of \($\)1,000 is made at a constant annual interest rate. The graphs above show the corresponding future value \(v\), in dollars, of the investment for different annual interest rates, \(r\), after 20 years. One graph shows the value when the interest is compounded daily, and the other graph shows the value when the interest is compounded annually. Which of the following statements is true?

- As \(r\) increases at a constant rate, \(v\) increases more rapidly if interest is compounded annually rather than daily.
- As \(r\) increases at a constant rate, \(v\) increases more rapidly if interest is compounded daily rather than annually.
- As \(r\) increases at a constant rate, the difference in interest compounded daily and interest compounded annually increases at a constant rate.
- If \(r\) = 15% and interest is compounded annually, a \($\)1,000 investment will be worth \($\)20,000 after 20 years.

**Answer/Explanation**

Ans: B

*Question*

\(P = 215(1.005)^\frac{t}{3}\)

The equation above can be used to model the population, in thousands, of a certain city \(t\) years after 2000. According to the model, the population is predicted to increase by 0.5% every n months. What is the value of n ?

- 3
- 4
- 12
- 36

**Answer/Explanation**

Ans: D

*Question*

\(P(t)=60(3)^\frac{t}{2}\)

The number of microscopic organisms in a petri dish grows exponentially with time. The function \(P\) above models the number of organisms after growing for \(t\) days in the petri dish. Based on the function, which of the following statements is true?

- The predicted number of organisms in the dish triples every two days.
- The predicted number of organisms in the dish doubles every three days.
- The predicted number of organisms in the dish triples every day.
- The predicted number of organisms in the dish doubles every day.

**Answer/Explanation**

Ans: A

*Question*

\(f(n) = 5.77(0.98^n)\)

The function above can be used to estimate the number of farms, \(f(n)\), in millions, in the United States for \(0 \leq n \leq 72\), where \(n\) is the number of years after 1940. Which of the following is the best interpretation of the number 5.77 in this context?

- The estimated number of farms, in millions, in 1940
- The estimated number of farms, in millions, \(n\) years after 1940
- The estimated decrease in the number of farms, in millions, each year after 1940
- The estimated percent by which the number of farms decreased from each year to the next after 1940

**Answer/Explanation**

Ans: A