7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Mini Exam 7\",\"description\":\"Timed practice exam\",\"url\":\"https://www.theschoolofmathematics.com/quiz/ap-precalculus-mini-exam-7\",\"timeRequired\":\"PT25M\",\"numberOfQuestions\":10,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Timed Mini Exams\"}}"}}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"BreadcrumbList\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https://www.theschoolofmathematics.com\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Quizzes\",\"item\":\"https://www.theschoolofmathematics.com/quiz\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Mini Exam 7\",\"item\":\"https://www.theschoolofmathematics.com/quiz/ap-precalculus-mini-exam-7\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","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":"Mini Exam 7"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","div",null,{"className":"text-lg md:text-xl text-gray-700 mb-8 leading-relaxed prose prose-lg max-w-none","dangerouslySetInnerHTML":{"__html":"

Timed practice exam

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Timed AP Precalculus Mini Exam 7

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

Timed practice exam

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":25,"niveau_difficulte":"$undefined","categorie":"Timed Mini Exams","nombre_questions":10,"score_minimum":80,"ordre_questions":"Aleatoire","questions":[{"id":"cmo52a23901bzjg683uy1aeue","texte_question":"$18","type_question":"QCM","explication":"

From the table, $$2,4,8,16,32$$

Each output is multiplied by $$2$$

So, the common ratio is $$b=2$$ Since $$b$$>$$1$$ the function shows exponential growth.

Final answer: $$\\text{B}$$

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

$$f$$ demonstrates exponential decay because $$0$$<$$b$$<$$1$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$f$$ demonstrates exponential growth because $$b$$>$$1$$.

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$f$$ is linear because the differences are constant.

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$f$$ is quadratic because second differences are constant.

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmo52a23901c4jg68vflydqe9","texte_question":"

The function $$f$$ is given by $$f(x)=x^2-4$$, and the function $$g$$ is given by $$g(x)=\\frac{x+2}{x}$$. Which of the following is an expression for $$f(g(x))$$?

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

Substitute

$$g(x)$$ into $$f(x)$$.

$$f(g(x))=(g(x))^2-4$$

Replace $$g(x)$$ with $$\\frac{x+2}{x}$$

Then $$f(g(x))=\\left(\\frac{x+2}{x}\\right)^2-4$$

Final answer: $$\\text{A}$$

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

$$\\left(\\frac{x+2}{x}\\right)^2-4$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$\\frac{x^2+4}{x^2}$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$\\frac{x^2}{x^2-4}$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$\\frac{x^2+2x-4}{x}$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmo52a23901c9jg68baskcq3l","texte_question":"

The function $$g$$ is given by $$g(x)=5\\sin(2x)$$. Which of the following is an equivalent form of $$g(x)$$?

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

Use the double-angle identity.

$$\\sin(2x)=2\\sin x\\cos x$$

Multiply by $$5$$.

$$g(x)=5\\sin(2x)=5(2\\sin x\\cos x)$$

$$g(x)=10\\sin x\\cos x$$

Final answer: $$\\text{A}$$

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

$$10\\sin x\\cos x$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$5\\cos^2x-5\\sin^2x$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$5-10\\sin^2x$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$5\\sin x\\cos x$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmo52a23901cejg68cwhzbctw","texte_question":"

A complex number has rectangular coordinates $$(-2,2\\sqrt{3})$$. Which of the following represents the number in polar form?

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

Find the modulus.

$$r=\\sqrt{(-2)^2+(2\\sqrt{3})^2}$$

$$r=\\sqrt{4+12}$$

$$r=4$$

The point is in Quadrant II.

Use the reference angle.

$$\\tan\\theta=\\frac{2\\sqrt{3}}{-2}=-\\sqrt{3}$$

So,

$$\\theta=\\frac{2\\pi}{3}$$

The polar form is:

$$4\\left(\\cos\\frac{2\\pi}{3}+i\\sin\\frac{2\\pi}{3}\\right)$$

Final answer: $$\\text{A}$$

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

$$4\\left(\\cos\\frac{2\\pi}{3}+i\\sin\\frac{2\\pi}{3}\\right)$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$2\\left(\\cos\\frac{\\pi}{3}+i\\sin\\frac{\\pi}{3}\\right)$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$4\\left(\\cos\\frac{\\pi}{3}+i\\sin\\frac{\\pi}{3}\\right)$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$2\\left(\\cos\\frac{2\\pi}{3}+i\\sin\\frac{2\\pi}{3}\\right)$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmo52a23901cjjg68mw2vaxoc","texte_question":"

The function $$f$$ is given by $$f(x)=2x^2-3x+1$$. Which of the following represents the graph of $$f$$ after a vertical dilation by a factor of $$3$$?

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

A vertical dilation by a factor of $$3$$ multiplies all outputs by $$3$$.

$$y=3f(x)$$

Substitute $$f(x)$$.

$$y=3(2x^2-3x+1)$$

Distribute.

$$y=6x^2-9x+3$$

Final answer: $$6x^2-9x+3$$

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

$$6x^2-9x+3$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$2(3x)^2-3(3x)+1$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$2x^2-3x+3$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$6x^2-3x+1$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmo52a23a01cojg68ak6ccvpc","texte_question":"

The functions $$f$$ and $$g$$ are defined for all real numbers such that $$g(x)=f(3(x+2))$$. Which of the following sequences of transformations maps the graph of $$f$$ to the graph of $$g$$ in the same $$xy$$-plane?

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

Rewrite the expression.

$$g(x)=f(3(x+2))$$

First apply the factor inside.

This gives a horizontal compression by a factor of $$\\frac{1}{3}$$.

Then the $$+2$$ shifts the graph left $$2$$ units.

Final answer: $$\\text{C}$$

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

A horizontal dilation by a factor of $$3$$, followed by a translation left $$2$$ units.

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

A horizontal dilation by a factor of $$\\frac{1}{3}$$, followed by a translation right $$2$$ units.

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

A horizontal dilation by a factor of $$\\frac{1}{3}$$, followed by a translation left $$2$$ units.

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

A horizontal dilation by a factor of $$3$$, followed by a translation right $$2$$ units.

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmo52a23a01ctjg68k7wkidng","texte_question":"

Let $$f$$ be a rational function. The numerator has a zero at $$x=2$$ but not at $$x=-3$$. The denominator has zeros at both $$x=2$$ and $$x=-3$$. The multiplicities at $$x=2$$ are equal in numerator and denominator. Which of the following is true?

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

At $$x=2$$, the zero cancels because multiplicities are equal.

This creates a hole.

At $$x=-3$$, the denominator is zero but the numerator is not.

This creates a vertical asymptote.

Final answer: $$\\text{C}$$

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

The graph has holes at both $$x=2$$ and $$x=-3$$.

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

The graph has a vertical asymptote at $$x=2$$ and a hole at $$x=-3$$.

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

The graph has a hole at $$x=2$$ and a vertical asymptote at $$x=-3$$.

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

The graph has vertical asymptotes at both $$x=2$$ and $$x=-3$$.

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmo52a23a01cyjg68etau1bzj","texte_question":"$19","type_question":"QCM","explication":"

From the table, $$5,10,20,40,80$$

Each value is multiplied by $$2$$

So, the model is exponential.

The function is $$y=5(2)^t$$

Final answer: $$5(2)^t$$

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

$$y=5+5t$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$y=5(2)^t$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$y=5(1.5)^t$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$y=5t^2$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmo52a23a01d3jg68hn1hg4g5","texte_question":"

The polynomial $$k(x)=ax^4+bx^3-6$$ has real coefficients. One zero is $$1+i$$. Another zero is $$2$$. Each zero has multiplicity $$1$$. Which of the following must be true?

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

The polynomial has real coefficients.

If $$1+i$$ is a zero, then its complex conjugate must also be a zero.

So, $$1-i$$ is also a zero.

Final answer: $$1-i\\text{ is also a zero}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1-i$$ is also a zero.","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

The graph has exactly two $$x$$-intercepts.

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

All zeros are real.

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

The polynomial has degree $$3$$.

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmo52a23b01d8jg682ziug8s8","texte_question":"

Consider the functions $$f(x)=\\log_5(x-2)+\\log_5(x+1)$$ and $$g(x)=\\log_5(x+3)$$. What are all $$x$$-values where the graphs intersect?

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

Set the functions equal.

$$\\log_5(x-2)+\\log_5(x+1)=\\log_5(x+3)$$

Use the product rule.

$$\\log_5\\left((x-2)(x+1)\\right)=\\log_5(x+3)$$

Set the arguments equal.

$$(x-2)(x+1)=x+3$$

Expand.

$$x^2-x-2=x+3$$

$$x^2-2x-5=0$$

Solve.

$$x=\\frac{2\\pm\\sqrt{24}}{2}$$

$$x=1\\pm\\sqrt{6}$$

Use the domain restrictions.

$$x-2$$>$$0$$

$$x+1$$>$$0$$

So,

$$x$$>$$2$$

Only

$$x=1+\\sqrt{6}$$ is valid.

Final answer: $$x=1+\\sqrt{6}$$

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

$$x=1$$ only

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$x=4$$ only

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$x=-3$$ and $$x=1$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$x=1+\\sqrt{6}$$ only

","correcte":true,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Timed-Mini-Exams"],"date":"2026-04-18T17:27:55.095Z","modified":"2026-04-19T01:02:25.556Z"}}]]}]}]]