7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Quiz 6\",\"description\":\"SAT Math Advanced Quizzes Quiz 6\",\"url\":\"https://theschoolofmathematics.com/quiz/sat-math-advanced-quizzes-quiz-6\",\"educationalLevel\":\"Advanced\",\"numberOfQuestions\":10,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Advanced Quizzes\"}}"}}],["$","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\":\"Quiz 6\",\"item\":\"https://theschoolofmathematics.com/quiz/sat-math-advanced-quizzes-quiz-6\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","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":"Quiz 6"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","div",null,{"className":"text-lg md:text-xl text-gray-700 mb-8 leading-relaxed prose prose-lg max-w-none","dangerouslySetInnerHTML":{"__html":"

SAT Math Advanced Quizzes Quiz 6

"}}],["$","div",null,{"className":"grid grid-cols-2 md:grid-cols-3 gap-4","children":["$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":"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"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":10}]]}]]}],["$","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":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M9 12l2 2 4-4m6 2a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Level"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"Advanced"}]]}]]}],["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","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"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"Advanced Quizzes"}]]}]]}]]}]]}]]}],["$","$L17",null,{"quiz":{"id":2360,"prismaId":"cmq0jyt1o01krjgodiozle3g1","slug":"sat-math-advanced-quizzes-quiz-6","title":{"rendered":"Quiz 6"},"content":{"rendered":"

SAT Math Advanced Quizzes - Quiz 6.

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

SAT Math Advanced Quizzes Quiz 6

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":"$undefined","niveau_difficulte":"Advanced","categorie":"Advanced Quizzes","nombre_questions":10,"score_minimum":70,"ordre_questions":"Fixe","questions":[{"id":"cmq0jyt1v01ktjgod4c1q4923","texte_question":"

If $$\\frac{\\sqrt[3]{x^2}}{\\sqrt[8]{x^5}}=x^{\\frac{a}{b}}$$ for all positive values of $$x$$, what is the value of $$\\frac{a}{b}$$?

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

Rewrite each radical using rational exponents.

$$\\sqrt[3]{x^2}=x^{\\frac{2}{3}}$$

$$\\sqrt[8]{x^5}=x^{\\frac{5}{8}}$$

Now divide powers with the same base.

$$\\frac{x^{\\frac{2}{3}}}{x^{\\frac{5}{8}}}=x^{\\frac{2}{3}-\\frac{5}{8}}$$

Find a common denominator.

$$\\frac{2}{3}=\\frac{16}{24}$$

$$\\frac{5}{8}=\\frac{15}{24}$$

Subtract.

$$\\frac{16}{24}-\\frac{15}{24}=\\frac{1}{24}$$

Therefore:

$$x^{\\frac{a}{b}}=x^{\\frac{1}{24}}$$

So:

$$\\frac{a}{b}=\\frac{1}{24}$$

Final answer: $$\\frac{1}{24}$$

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

$$\\frac{1}{24}$$

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

$$\\frac{1}{4}$$

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

$$\\frac{3}{25}$$

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

$$\\frac{2}{3}$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyt1z01kzjgodk7ch6im9","texte_question":"

The function $$p$$ is defined by $$p(x)=a((x+6)^2-b)((x+6)^2-c)$$ where $$a$$, $$b$$, and $$c$$ are constants. In the $$xy$$-plane, the graph of $$y=p(x)$$ passes through the points $$(-7,49)$$ and $$(0,784)$$. What is the value of $$p(-12)+p(-5)$$?

","type_question":"TexteLibre","explication":"

The function depends on $$(x+6)^2$$.

So, any two inputs with the same value of $$(x+6)^2$$ have the same output.

Compare $$x=-7$$ and $$x=-5$$.

For $$x=-7$$:

$$(-7+6)^2=(-1)^2=1$$

For $$x=-5$$:

$$(-5+6)^2=(1)^2=1$$

Therefore:

$$p(-5)=p(-7)$$

Since the graph passes through $$(-7,49)$$:

$$p(-7)=49$$

So:

$$p(-5)=49$$

Now compare $$x=0$$ and $$x=-12$$.

For $$x=0$$:

$$(0+6)^2=6^2=36$$

For $$x=-12$$:

$$(-12+6)^2=(-6)^2=36$$

Therefore:

$$p(-12)=p(0)$$

Since the graph passes through $$(0,784)$$:

$$p(0)=784$$

So:

$$p(-12)=784$$

Now add.

$$p(-12)+p(-5)=784+49$$

$$=833$$

Final answer: $$833$$

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

833

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyt2301l2jgodn2jcdwem","texte_question":"

$$2\\cos(90-a)^\\circ\\cdot\\cos b^\\circ+\\sin(a+y)^\\circ\\cdot\\sin(90-b)^\\circ$$

In the given expression, $$a$$ and $$b$$ are constants such that $$\\sin a^\\circ=0.50$$, $$\\cos b^\\circ=0.99$$, and $$0$$ < $$b$$ < $$90$$. What is the value of the expression when $$y=0$$?

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

Substitute $$y=0$$ into the expression.

$$2\\cos(90-a)^\\circ\\cos b^\\circ+\\sin a^\\circ\\sin(90-b)^\\circ$$

Use cofunction identities. $$\\cos(90-a)^\\circ=\\sin a^\\circ$$

$$\\sin(90-b)^\\circ=\\cos b^\\circ$$

So, the expression becomes:

$$2\\sin a^\\circ\\cos b^\\circ+\\sin a^\\circ\\cos b^\\circ$$

Combine like terms.

$$3\\sin a^\\circ\\cos b^\\circ$$

Substitute the given values.

$$3(0.50)(0.99)$$

$$=1.485$$

Final answer: $$1.485$$

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

$$1.758$$

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

$$1.485$$

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

$$2.325$$

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

$$2.675$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyt2e01l8jgodcxzr7yez","texte_question":"

Function $$f$$ is defined by $$f(x)=a^x+b$$, where $$a$$ and $$b$$ are constants. In the $$xy$$-plane, the graph of $$y=f(x)-11$$ has a $$y$$-intercept at $$(0,-\\frac{29}{4})$$. The product of $$a$$ and $$b$$ is $$\\frac{66}{4}$$. What is the value of $$a$$?

","type_question":"TexteLibre","explication":"

The graph of $$y=f(x)-11$$ has a $$y$$-intercept at $$(0,-\\frac{29}{4})$$.

So, when $$x=0$$:

$$f(0)-11=-\\frac{29}{4}$$

Add $$11$$ to both sides.

$$f(0)=11-\\frac{29}{4}$$

$$f(0)=\\frac{44}{4}-\\frac{29}{4}$$

$$f(0)=\\frac{15}{4}$$

Since:

$$f(x)=a^x+b$$

we have:

$$f(0)=a^0+b$$

$$f(0)=1+b$$

So:

$$1+b=\\frac{15}{4}$$

$$b=\\frac{11}{4}$$

The product of $$a$$ and $$b$$ is:

$$ab=\\frac{66}{4}$$

Substitute

$$b=\\frac{11}{4}$$.

$$a\\left(\\frac{11}{4}\\right)=\\frac{66}{4}$$

Multiply both sides by $$\\frac{4}{11}$$.

$$a=6$$

Final answer: $$6$$

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

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyt2i01lbjgodfqz1jdrq","texte_question":"

Consider the system

$$2x+3y=7$$

$$10x+15y=35$$

For each real number $$r$$, which of the following points lies on the graph of each equation in the $$xy$$-plane for the given system?

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

The second equation is a multiple of the first equation.

$$10x+15y=35$$

Divide by $$5$$.

$$2x+3y=7$$

So, both equations represent the same line.

Solve the first equation for $$x$$ in terms of $$y$$.

$$2x+3y=7$$

Let: $$y=r$$

Then:

$$2x+3r=7$$

$$2x=7-3r$$

$$x=\\frac{7-3r}{2}$$

$$x=-\\frac{3r}{2}+\\frac{7}{2}$$

So, a point on both graphs for each real number $$r$$ is:

$$\\left(-\\frac{3r}{2}+\\frac{7}{2},r\\right)$$

Final answer: $$\\left(-\\frac{3r}{2}+\\frac{7}{2},r\\right)$$

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

$$\\left(\\frac{r}{5}+7,-\\frac{r}{5}+35\\right)$$

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

$$\\left(-\\frac{3r}{2}+\\frac{7}{2},r\\right)$$

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

$$\\left(r,\\frac{2r}{3}+\\frac{7}{3}\\right)$$

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

$$\\left(r,-\\frac{3r}{2}+\\frac{7}{2}\\right)$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyt2m01lhjgods2ymgopi","texte_question":"

In the equation $$(3x+p)(5x^2-45)(2x^2-16x+6p)=0$$ $$p$$ is a positive constant. The sum of the solutions to the equation is $$\\frac{20}{3}$$. What is the value of $$p$$?

","type_question":"TexteLibre","explication":"

The equation is already factored. The solutions come from each factor equal to $$0$$.

First factor:

$$3x+p=0$$

$$x=-\\frac{p}{3}$$

Second factor:

$$5x^2-45=0$$

$$5(x^2-9)=0$$

$$x^2=9$$

$$x=3$$ or $$x=-3$$

The sum of these two solutions is:

$$3+(-3)=0$$

Third factor:

$$2x^2-16x+6p=0$$

For a quadratic $$ax^2+bx+c=0$$, the sum of the solutions is: $$-\\frac{b}{a}$$

Here: $$a=2$$, $$b=-16$$

So, the sum of the solutions from this factor is:

$$-\\frac{-16}{2}=8$$

Therefore, the total sum of all solutions is: $$-\\frac{p}{3}+0+8$$

This equals $$\\frac{20}{3}$$.

$$8-\\frac{p}{3}=\\frac{20}{3}$$

Rewrite $$8$$ as $$\\frac{24}{3}$$.

$$\\frac{24}{3}-\\frac{p}{3}=\\frac{20}{3}$$

$$24-p=20$$

$$p=4$$

Final answer: $$4$$

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

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyt2p01lkjgodz2enznsq","texte_question":"

The given polynomial $$54x^4+216x^2+210$$ has factors in the form $$k(ax^2+b)(cx^2+d)$$ If $$a$$, $$b$$, $$c$$, $$d$$, and $$k$$ are integers, what is the smallest possible value of $$ab$$?

","type_question":"TexteLibre","explication":"

Factor out the greatest common factor.

$$54x^4+216x^2+210=6(9x^4+36x^2+35)$$

Now factor the quadratic in $$x^2$$.

Let: $$u=x^2$$

Then:

$$9u^2+36u+35$$

Factor.

$$9u^2+36u+35=(3u+5)(3u+7)$$

Substitute back $$u=x^2$$.

$$9x^4+36x^2+35=(3x^2+5)(3x^2+7)$$

So:

$$54x^4+216x^2+210=6(3x^2+5)(3x^2+7)$$

One valid assignment is:

$$k=6$$, $$a=3$$, $$b=5$$

Then: $$ab=15$$

But the factors can also be written by moving a negative sign into one factor and another negative sign into $$k$$.

For example:

$$6(3x^2+5)(3x^2+7)=6(-3x^2-5)(-3x^2-7)$$

Then:

$$a=-3$$ $$b=-5$$ and $$ab=15$$

The smallest positive value is $$15$$.

Final answer: $$15$$

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

","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyt2t01lnjgodsfsxzd3r","texte_question":"

For the function $$f$$, for each increase in the value of $$x$$ by $$c$$, where $$c$$ is a positive constant, the value of $$f(x)$$ increases by a factor of $$27$$. Which of the following equivalent forms of the function $$f$$ displays $$\\frac{1}{c}$$ as a coefficient of $$x$$?

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

If the value of $$f(x)$$ increases by a factor of $$27$$ every time $$x$$ increases by $$c$$, then the function can be written in the form:

$$f(x)=48(27)^{\\frac{x}{c}}$$

Since:

$$\\frac{x}{c}=\\frac{1}{c}x$$ this form displays $$\\frac{1}{c}$$ as the coefficient of $$x$$.

Now rewrite $$27$$ as a power of $$3$$.

$$27=3^3$$

So:

$$48(27)^{\\frac{1}{c}x}=48(3^3)^{\\frac{1}{c}x}$$

Final answer: $$f(x)=48(3^3)^{\\frac{1}{c}x}$$

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

$$f(x)=48(3)^{\\frac{1}{2}x}$$

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

$$f(x)=48(3^3)^{\\frac{1}{c}x}$$

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

$$f(x)=48(9)^{\\frac{1}{4}x}$$

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

$$f(x)=48\\left(27^{\\frac{1}{3}x}\\right)^{\\frac{1}{2}}$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyt2x01ltjgod8ckd1vc5","texte_question":"

In the given equation, $$57x^2+(57b+a)x+ab=0$$ where $$a$$ and $$b$$ are positive constants. The product of the solutions to the given equation is $$kab$$, where $$k$$ is a constant. What is the value of $$k$$?

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

For a quadratic equation:

$$Ax^2+Bx+C=0$$

The product of the solutions is: $$\\frac{C}{A}$$

Here: $$A=57$$ and $$C=ab$$

So, the product of the solutions is: $$\\frac{ab}{57}$$

The problem says the product is: $$kab$$

So:

$$kab=\\frac{ab}{57}$$

Divide both sides by $$ab$$.

$$k=\\frac{1}{57}$$

Final answer: $$\\frac{1}{57}$$

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

$$\\frac{1}{57}$$

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

$$\\frac{1}{19}$$

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

$$1$$

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

$$57$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmq0jyt3701lzjgod68bxxmc4","texte_question":"$18","type_question":"QCM","explication":"

Use the unit circle.

The value of $$\\sin\\theta$$ represents the $$y$$-coordinate.

We are given:

$$\\sin\\theta=\\frac{\\sqrt{3}}{2}$$

On the unit circle, this happens at angles with reference angle $$60^\\circ$$.

The figure shows that $$\\theta$$ is in Quadrant II.

In Quadrant II, cosine is negative.

For a $$60^\\circ$$ reference angle:

$$\\cos60^\\circ=\\frac{1}{2}$$

Since the angle is in Quadrant II:

$$\\cos\\theta=-\\frac{1}{2}$$

Final answer: $$-\\frac{1}{2}$$

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

$$\\frac{\\sqrt{3}}{2}$$

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

$$\\frac{1}{2}$$

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

$$-\\frac{1}{2}$$

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

$$-\\frac{\\sqrt{3}}{2}$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Advanced Quizzes"],"date":"2026-06-05T06:38:07.501Z","modified":"2026-06-05T06:38:07.501Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L19",null,{"targetType":"quiz","targetSlug":"sat-math-advanced-quizzes-quiz-6"}]}]]}]}]]

Quiz 6

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