Section I Part B
Section I Part B practice
Loading quiz...
Frequently Asked Questions
Quick answers based on this quiz content.
Solve the logarithmic inequality $$\log_2(x-5)+\log_2(x-1)\le3$$⌄
Combine logarithms.$$\log_2((x-5)(x-1))\le3.$$Rewrite in exponential form.$$(x-5)(x-1)\le2^3.$$$$(x-5)(x-1)\le8.$$Expand.$$x^2-6x+5\le8.$$$$x^2-6x-3\le0.$$Solve the quadratic.$$x=\frac{6\pm\sqrt{36+12}}{2}$$$$x=\frac{6\pm\sqrt{48}}{2}$$$$x=\frac{6\pm4\sqrt{3}}{2}$$$$x=3\pm2\sqrt{3}.$$So the quadratic inequality gives:$$3-2\sqrt{3}\le x\le3+2\sqrt{3}.$$Apply the domain.Since:$$(x-5)(x-1)$$>$$0,$$we need:$$x$$<$$1$$ or $$x$$>$$5.$$Intersect this with:$$3-2\sqrt{3}\le x\le3+2\sqrt{3}.$$Because$$3-2\sqrt{3}$$<$$1$$ and $$3+2\sqrt{3}$$>$$5,$$the valid solution is:$$[3-2\sqrt{3},1)\cup(5,3+2\sqrt{3}].$$Final answer: $$[3-2\sqrt{3},1)\cup(5,3+2\sqrt{3}]$$
The function $$f(x)=3^{-\frac{x}{3}}\sin\left(\frac{\pi x}{3}\right)$$ represents a sound wave. What is the average rate of change of $$f(x)$$ over the interval $$[-3,3]$$ to the nearest hundredth?⌄
Use average rate of change.$$\frac{f(3)-f(-3)}{3-(-3)}.$$Compute values.$$f(3)=3^{-1}\sin(\pi)=0$$$$f(-3)=3^{1}\sin(-\pi)=0.$$So:$$\frac{0-0}{6}=0$$Final answer: $$0$$
At what value of $$x$$ does the absolute minimum of $$g(x)=e^{0.8x}\cos(x)$$ occur over the interval $$[0,2\pi]$$?⌄
Test key points.$$g(0)=1$$$$g(\pi)=e^{0.8\pi}(-1)$$$$g(2\pi)=e^{1.6\pi}(1).$$Since $$e^{0.8\pi}$$ is positive and multiplied by $$-1,$$$$g(\pi)$$ is the smallest.Final answer: $$\pi$$
What is the solution to $$x^3-6x^2-15x+54$$>$$0$$?⌄
Factor the polynomial.$$x^3-6x^2-15x+54=(x-3)(x-6)(x+3).$$Critical points:$$x=-3,3,6.$$Test intervals.For $$x$$<$$-3,$$ the expression is negative.For $$-3$$<$$x$$<$$3,$$ the expression is positive.For $$3$$<$$x$$<$$6,$$ the expression is negative.For $$x$$>$$6,$$ the expression is positive.So the polynomial is positive on:$$(-3,3)\cup(6,\infty).$$Final answer: $$(-3,3)\cup(6,\infty)$$
If $$y=\frac{ax+b}{cx+d}$$ has a horizontal asymptote $$y=3$$ and a vertical asymptote at $$x=-2$$, and $$c=2$$, what is the value of $$a+d$$?⌄
Horizontal asymptote:$$\frac{a}{c}=3$$$$\frac{a}{2}=3$$$$a=6.$$Vertical asymptote:$$cx+d=0$$$$2x+d=0.$$At $$x=-2$$:$$2(-2)+d=0$$$$-4+d=0$$$$d=4.$$Compute:$$a+d=6+4=10.$$Final answer: $$10$$
The profit for a company is modeled by $$P(t)=12t^2+18t+450,$$ where $$t$$ is the number of years since $$2005$$ and $$P(t)$$ is in thousands of dollars. If this trend continues, in what year will the company's profit be $$600{,}000$$?⌄
Set the profit equal to $$600$$ because the model is in thousands of dollars.$$12t^2+18t+450=600.$$Move all terms to one side.$$12t^2+18t-150=0.$$Divide by $$6.$$$$2t^2+3t-25=0.$$Use the quadratic formula.$$t=\frac{-3\pm\sqrt{3^2-4(2)(-25)}}{2(2)}$$$$t=\frac{-3\pm\sqrt{9+200}}{4}$$$$t=\frac{-3\pm\sqrt{209}}{4}.$$Take the positive solution.$$t\approx\frac{-3+14.46}{4}$$$$t\approx2.86.$$This is about $$2.86$$ years after $$2005.$$So, the year is about: $$2008.$$Final answer: $$2008$$
A radioactive substance has a half-life of $$40$$ days. If an experiment begins with $$250$$ grams of the substance, which equation approximates the amount $$A$$ of the substance after $$t$$ days?⌄
Use the half-life model.$$A=250\left(\frac{1}{2}\right)^{\frac{t}{40}}.$$Rewrite the exponent.$$A=250\left(\left(\frac{1}{2}\right)^{\frac{1}{40}}\right)^t.$$Approximate the base.$$\left(\frac{1}{2}\right)^{\frac{1}{40}}\approx0.982821.$$So the model can be written as:$$A=250(0.982821)^t.$$Final answer: $$A=250(0.982821)^t$$
The heights of consecutive bounces of a ball form a geometric sequence. The height of the second bounce is $$96$$ inches, and the height of the fourth bounce is $$24$$ inches. To the nearest inch, what is the height of the sixth bounce?⌄
Use the geometric sequence formula.$$a_n=a_1r^{n-1}.$$From the problem:$$a_2=96$$$$a_4=24.$$So:$$\frac{a_4}{a_2}=r^2.$$$$\frac{24}{96}=r^2.$$$$\frac{1}{4}=r^2.$$Since bounce heights are positive and decreasing:$$r=\frac{1}{2}.$$Now move from the fourth bounce to the sixth bounce.$$a_6=a_4r^2.$$$$a_6=24\left(\frac{1}{2}\right)^2.$$$$a_6=24\cdot\frac{1}{4}.$$$$a_6=6.$$Final answer: $$6$$