Mini Exam 2 — Timed Mini Exams
Practice Mini Exam 2 — Timed Mini Exams on The School of Mathematics (Timed Mini Exams) with 10 scored questions in about 15 min. This set covers problems such as: “The velocity of a stone that is thrown upward is represented by the equation v=-4t^2-16t+80 , where v repre…”; “If x=3y+4 and y=-x-2 , what is the value of xy ?”; “If 2^{x+3}=8^2 , what is the value of x ?”. Work carefully, check explanations for misses, then retake to track improvement before test day.
Questions in this quiz (10)
The velocity of a stone that is thrown upward is represented by the equation $$v=-4t^2-16t+80$$, where $$v$$ represents velocity and $$t$$ represents time, in seconds. After how many seconds does the velocity of the stone become zero?
- $$2.9$$
- $$4.1$$
- $$1.6$$
- $$2.4$$
If $$x=3y+4$$ and $$y=-x-2$$, what is the value of $$xy$$?
- $$\frac{3}{2}$$
- $$\frac{1}{4}$$
- $$\frac{1}{2}$$
- $$\frac{3}{4}$$
If $$2^{x+3}=8^2$$, what is the value of $$x$$?
- 3
- 1
- 4
- 2
If $$\sqrt{8}\times\sqrt{m}=4b$$, what is the value of $$m$$ in terms of $$b$$?
- $$2b$$
- $$2b^2$$
- $$4b^2$$
- $$8b^2$$
If $$f(x)=3x^2+c$$ and $$g(x)=x+1$$, where $$c$$ is a constant, and $$\frac{f(g(2))}{f(1)}=2,$$ what is the value of $$c$$?
- $$24$$
- $$27$$
- $$25$$
- $$21$$
Depreciation is the decrease in value of an item. Four years ago, Emily bought a car for 16000. The car is now worth 10000. Assuming that the value of the car depreciates linearly, what is its yearly depreciation?
- $$1500$$
- $$1000$$
- $$1200$$
- $$1400$$
If $$0$$<$$a$$<$$2$$<$$b$$<$$c$$, each of the following could be less than $$\frac{1}{2}$$ EXCEPT:
- $$\frac{a}{b}$$
- $$\frac{a}{2}$$
- $$\frac{2}{b}$$
- $$\frac{c}{a}$$
The graph shows a company's Revenue and Costs (in thousands of dollars) from January to May. For which month was profit greatest?(Profit is defined as: $$\text{Profit}=\text{Revenue}-\text{Costs}.$$)
- January
- March
- April
- May
$$4,12,5,9,10$$The mean of the list of numbers above is what fraction of the sum of the five numbers?
- $$\frac{5}{8}$$
- $$\frac{8}{5}$$
- $$\frac{1}{5}$$
- $$\frac{4}{9}$$
Circle $$P$$ has a radius of $$5k$$, and circle $$Q$$ has a radius of $$40k-5$$, where $$k$$ is a positive constant. The area of circle $$Q$$ is how many times the area of circle $$P$$?
- $$\frac{1}{5k^2}$$
- $$\frac{(40k-5)^2}{25k^2}$$
- $$\frac{40}{k^2}$$
- $$\frac{(40k)^2}{25}$$