7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Section I Part B\",\"description\":\"Section I Part B practice\",\"url\":\"https://theschoolofmathematics.com/quiz/ap-precalculus-full-practice-exam-3-section-1-part-b\",\"timeRequired\":\"PT40M\",\"numberOfQuestions\":11,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Full Practice Exam 3\"}}"}}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"BreadcrumbList\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https://theschoolofmathematics.com\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Quizzes\",\"item\":\"https://theschoolofmathematics.com/quiz\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Section I Part B\",\"item\":\"https://theschoolofmathematics.com/quiz/ap-precalculus-full-practice-exam-3-section-1-part-b\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","div",null,{"className":"container mx-auto px-4 py-8 md:py-12","children":[["$","div",null,{"className":"mb-12","children":["$undefined",["$","div",null,{"className":"mb-8","children":["$","h1",null,{"className":"text-4xl md:text-5xl lg:text-6xl font-bold text-gray-900 mb-4","children":"Section I Part B"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","div",null,{"className":"text-lg md:text-xl text-gray-700 mb-8 leading-relaxed prose prose-lg max-w-none","dangerouslySetInnerHTML":{"__html":"

Section I Part B practice

"}}],["$","div",null,{"className":"grid grid-cols-2 md:grid-cols-3 gap-4","children":[["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M12 8v4l3 3m6-3a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Duration"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"40 min"}]]}]]}],["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M8.228 9c.549-1.165 2.03-2 3.772-2 2.21 0 4 1.343 4 3 0 1.4-1.278 2.575-3.006 2.907-.542.104-.994.54-.994 1.093m0 3h.01M21 12a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Questions"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":11}]]}]]}],"$undefined",["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M7 7h.01M7 3h5c.512 0 1.024.195 1.414.586l7 7a2 2 0 010 2.828l-7 7a2 2 0 01-2.828 0l-7-7A1.994 1.994 0 013 12V7a4 4 0 014-4z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Category"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"Full Practice Exam 3"}]]}]]}]]}]]}]]}],["$","$L17",null,{"quiz":{"id":2301,"prismaId":"cmpz9qsbs04iajgllcf309ot2","slug":"ap-precalculus-full-practice-exam-3-section-1-part-b","title":{"rendered":"Section I Part B"},"content":{"rendered":"

Multiple choice section for Full Practice Exam 3

"},"excerpt":{"rendered":"

Section I Part B practice

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":40,"niveau_difficulte":"$undefined","categorie":"Full Practice Exam 3","nombre_questions":11,"score_minimum":80,"ordre_questions":"Aleatoire","questions":[{"id":"cmpz9qsbv04icjgllialah6t5","texte_question":"

Solve the logarithmic inequality $$\\log_2(x-5)+\\log_2(x-1)\\le3$$

","type_question":"QCM","explication":"

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: $$[3-2\\sqrt{3},1)\\cup(5,3+2\\sqrt{3}]$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$[3-2\\sqrt{3},1)\\cup(5,3+2\\sqrt{3}]$$

","correcte":true,"explication":"Correct solution.","imageUrl":"$undefined"},{"texte":"$$$1Endpoint should be included.

","imageUrl":"$undefined"},{"texte":"

$$x\\le3+2\\sqrt{3}$$

","correcte":false,"explication":"

Domain restriction missing.

","imageUrl":"$undefined"}]},{"id":"cmpz9qsby04iijgll0bpvwpau","texte_question":"

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?

","type_question":"QCM","explication":"

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: $$0$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-0.53$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"

$$0$$

","correcte":true,"explication":"Correct average rate.","imageUrl":"$undefined"},{"texte":"$$$0.48$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"},{"texte":"

$$0.53$$

","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9qsc104iojgllxbvlpjs1","texte_question":"

At what value of $$x$$ does the absolute minimum of $$g(x)=e^{0.8x}\\cos(x)$$ occur over the interval $$[0,2\\pi]$$?

","type_question":"QCM","explication":"

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: $$\\pi$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"Positive value.","imageUrl":"$undefined"},{"texte":"$$$\\pi$$","correcte":true,"explication":"Gives largest negative value.","imageUrl":"$undefined"},{"texte":"$$$\\frac{\\pi}{2}$$","correcte":false,"explication":"

Not minimum.

","imageUrl":"$undefined"},{"texte":"

$$2\\pi$$

","correcte":false,"explication":"

Large positive value.

","imageUrl":"$undefined"}]},{"id":"cmpz9qsc504iujgll5yduxrnx","texte_question":"

What is the solution to $$x^3-6x^2-15x+54$$>$$0$$?

","type_question":"QCM","explication":"

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: $$(-3,3)\\cup(6,\\infty)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(-3,3)\\cup(6,\\infty)$$","correcte":true,"explication":"Correct sign analysis.","imageUrl":"$undefined"},{"texte":"$$$(-\\infty,-3)\\cup(3,6)$$","correcte":false,"explication":"Wrong intervals.","imageUrl":"$undefined"},{"texte":"$$$(-3,3)$$","correcte":false,"explication":"

Missing interval.

","imageUrl":"$undefined"},{"texte":"

$$(3,6)$$

","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9qsc904j0jgll5dbiuuam","texte_question":"

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 $$a+d$$?

","type_question":"QCM","explication":"

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: $$10$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$5$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$8$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"},{"texte":"

$$10$$

","correcte":true,"explication":"

Correct value.

","imageUrl":"$undefined"}]},{"id":"cmpz9qscb04j6jgllotrs7yla","texte_question":"

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 $$600{,}000$$?

","type_question":"QCM","explication":"

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: $$2008$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2007$$","correcte":false,"explication":"This is too early.","imageUrl":"$undefined"},{"texte":"$$$2008$$","correcte":true,"explication":"Correct year.","imageUrl":"$undefined"},{"texte":"$$$2009$$","correcte":false,"explication":"

This is too late.

","imageUrl":"$undefined"},{"texte":"

$$2010$$

","correcte":false,"explication":"

This is too late.

","imageUrl":"$undefined"}]},{"id":"cmpz9qsce04jcjgllt3jnv1aj","texte_question":"

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?

","type_question":"QCM","explication":"

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: $$A=250(0.982821)^t$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$A=250(0.98)^t$$

","correcte":true,"explication":"Correct half-life model.","imageUrl":"$undefined"},{"texte":"$$$A=250(0.5)^{40t}$$","correcte":false,"explication":"The exponent is incorrect.","imageUrl":"$undefined"},{"texte":"$$$A=250\\left(\\frac{1}{80}\\right)^t$$","correcte":false,"explication":"

This is not the half-life decay factor.

","imageUrl":"$undefined"},{"texte":"

$$A=250(1.017179)^t$$

","correcte":false,"explication":"

This represents growth, not decay.

","imageUrl":"$undefined"}]},{"id":"cmpz9qsch04jijgll5x6n8r81","texte_question":"

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?

","type_question":"QCM","explication":"

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: $$6$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$12$$","correcte":false,"explication":"This is too large.","imageUrl":"$undefined"},{"texte":"$$$10$$","correcte":false,"explication":"This does not follow the common ratio.","imageUrl":"$undefined"},{"texte":"$$$8$$","correcte":false,"explication":"

This is incorrect.

","imageUrl":"$undefined"},{"texte":"

$$6$$

","correcte":true,"explication":"

Correct sixth bounce height.

","imageUrl":"$undefined"}]},{"id":"cmpz9qsck04jojgll4iysq0i0","texte_question":"$18","type_question":"QCM","explication":"

Test the model at $$p=1$$ because:

$$\\ln(1)=0.$$

For the equation:

$$h=-8.12\\ln p+0.05$$

we get:

$$h=-8.12\\ln(1)+0.05$$

$$h=0.05.$$

This is very close to the table value $$0.$$

Now test $$p=0.61.$$

$$h=-8.12\\ln(0.61)+0.05$$

$$h\\approx-8.12(-0.494)+0.05$$

$$h\\approx4.06.$$

This is very close to the table value $$4.$$

So the best model is:

$$h=-8.12\\ln p+0.05.$$

Final answer: $$h=-8.12\\ln p+0.05$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$h=-8.12\\ln p+0.05$$","correcte":true,"explication":"Correct logarithmic model.","imageUrl":"$undefined"},{"texte":"$$$h=0.05-8.12p$$","correcte":false,"explication":"This is linear, not logarithmic.","imageUrl":"$undefined"},{"texte":"$$$h=8.12\\ln p+0.05$$","correcte":false,"explication":"

The sign is incorrect.

","imageUrl":"$undefined"},{"texte":"

$$h=-20.05+8.12\\ln p$$

","correcte":false,"explication":"

This does not fit the data pattern.

","imageUrl":"$undefined"}]},{"id":"cmpz9qscn04jujgll3tv6xg3g","texte_question":"

The function $$S(d)=450\\log d+120$$ relates $$S(d)$$, the speed of a storm in miles per hour, to $$d$$, the distance it travels in miles. If the speed reaches $$570$$ mph, estimate the distance the storm can travel.

","type_question":"QCM","explication":"

Set the speed equal to $$570.$$

$$450\\log d+120=570.$$

Subtract $$120.$$

$$450\\log d=450.$$

Divide by $$450.$$

$$\\log d=1.$$

Rewrite in exponential form.

$$d=10^1.$$

$$d=10.$$

Final answer: $$10$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$3\\text{ miles}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$10\\text{ miles}$$","correcte":true,"explication":"Correct distance.","imageUrl":"$undefined"},{"texte":"$$$30\\text{ miles}$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"},{"texte":"

$$300\\text{ miles}$$

","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9qscq04k0jglliy5bseg3","texte_question":"

If $$\\cos\\theta=\\frac{5}{13}$$ and $$0\\le\\theta\\le\\frac{\\pi}{2},$$ evaluate $$\\sin\\theta\\tan\\theta.$$

","type_question":"QCM","explication":"

Use a right triangle.

If $$\\cos\\theta=\\frac{5}{13},$$ then the adjacent side is $$5$$ and the hypotenuse is $$13.$$

Find the opposite side.

$$\\sqrt{13^2-5^2}$$

$$=\\sqrt{169-25}$$

$$=\\sqrt{144}$$

$$=12.$$

So:

$$\\sin\\theta=\\frac{12}{13}$$

and

$$\\tan\\theta=\\frac{12}{5}.$$

Now multiply.

$$\\sin\\theta\\tan\\theta=\\frac{12}{13}\\cdot\\frac{12}{5}$$

$$=\\frac{144}{65}.$$

Final answer: $$\\frac{144}{65}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{144}{65}$$","correcte":true,"explication":"Correct value.","imageUrl":"$undefined"},{"texte":"

$$\\frac{144}{169}$$

","correcte":false,"explication":"

This is $$\\sin^2\\theta$$.

","imageUrl":"$undefined"},{"texte":"$$$\\frac{12}{13}$$","correcte":false,"explication":"

This is only $$\\sin\\theta$$.

","imageUrl":"$undefined"},{"texte":"

$$\\frac{5}{12}$$

","correcte":false,"explication":"

Incorrect reciprocal.

","imageUrl":"$undefined"}]}]},"categories":["full-practice-exam-3"],"date":"2026-06-04T09:04:10.984Z","modified":"2026-06-04T09:04:10.984Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L19",null,{"targetType":"quiz","targetSlug":"ap-precalculus-full-practice-exam-3-section-1-part-b"}]}]]}]}]]

Section I Part B

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