Skip to main content
Logo image

Intermediate Algebra Functions and Graphs

Katherine Yoshiwara

Section A.3 Chapter 3 Quadratic Models

Subsection A.3.1 Extraction of Roots

Subsubsection A.3.1.1 Evaluate quadratic expressions

When squaring a negative number, don’t forget to enclose it in parentheses. For example, if \(x=-4\text{,}\) then
\begin{equation*} x^2=(-4)^2 = (-4)(-4) = 16 \end{equation*}
If we write \(-4^2\text{,}\) then only the 4 is squared, so we have
\begin{equation*} -4^2 = - (4)(4) = -16 \end{equation*}
🔗
Example A.3.1.
Evaluate for \(x=-6\text{.}\)
  1. \(\displaystyle 2x^2\)
    🔗
  2. \(\displaystyle 2-x^2\)
    🔗
  3. \(\displaystyle (2x)^2\)
    🔗
  4. \(\displaystyle (2-x)^2\)
    🔗
🔗
Solution.
Enclose \(-6\) in parentheses, and follow the order of operations.
  1. Square first, then multiply by 2: \(~~2x^2=2(-6)^2=2(36)=72\)
    🔗
  2. Square first, then subtract from 2: \(~~2-x^2=2-(-6)^2=2-36=-34\)
    🔗
  3. Multiply by 2 first, then square: \(~~(2x)^2=[2(-6)]^2=[-12]^2=144\)
    🔗
  4. Subtract from 2 first, then square: \(~~(2-x)^2=[2-(-6)]^2=[8]^2=64\)
    🔗
🔗
🔗
🔗
Example A.3.2.
Make a table of values for \(y=x^2+2x-8\text{,}\) and graph the equation.
🔗
Solution.
We plot the points from the table and connect them with a smooth curve.
🔗
\(x\) \(y\) \(\hphantom{0000}\)
\(-5\) \(7\) \((-5)^2+2(-5)-8=25-10-8\)
\(-4\) \(0\) \((-4)^2+2(-4)-8=16-8-8\)
\(-3\) \(-5\) \((-3)^2+2(-3)-8=9-6-8\)
\(-2\) \(-8\) \((-2)^2+2(-2)-8=4-4-8\)
\(-1\) \(-9\) \((-1)^2+2(-1)-8=1-2-8\)
\(0\) \(-8\) \((0)^2+2(0)-8=0+0-8\)
\(1\) \(-5\) \((1)^2+2(1)-8=1+2-8\)
\(2\) \(0\) \((2)^2+2(2)-8=4+4-8\)
\(3\) \(7\) \((3)^2+2(3)-8=9+6-8\)
parabola y=xsqd+2x-8
🔗
🔗
Checkpoint A.3.6.
Make a table of values for \(y=x^2-x-6\text{,}\) and graph the equation.
🔗
\(x\) \(y\)
\(-3\) \(\hphantom{0000}\)
\(-2\) \(\hphantom{0000}\)
\(-1\) \(\hphantom{0000}\)
\(0\) \(\hphantom{0000}\)
\(1\) \(\hphantom{0000}\)
\(2\) \(\hphantom{0000}\)
\(3\) \(\hphantom{0000}\)
\(4\) \(\hphantom{0000}\)
\(5\) \(\hphantom{0000}\)
grid
Answer.
\(x\) \(y\)
\(-3\) \(6\)
\(-2\) \(0\)
\(-1\) \(-4\)
\(0\) \(-6\)
\(1\) \(-6\)
\(2\) \(-4\)
\(3\) \(0\)
\(4\) \(6\)
\(5\) \(14\)
parabola
🔗
🔗
🔗

Subsubsection A.3.1.2 Use square roots

The simplest quadratic equations have the form
\begin{equation*} ax^2+c=0 \end{equation*}
They are missing a linear term, \(bx\text{,}\) and are not difficult to solve.
🔗
Example A.3.7.
Solve \(~3y^2-40=35~\) by extraction of roots.
🔗
Solution.
We isolate the quadratic term, \(y^2\text{,}\) and then extract roots.
\begin{align*} 3y^2-40 \amp = 35 \amp \amp \blert{\text{Add 40 to both sides.}}\\ 3y^2 \amp = 75 \amp \amp \blert{\text{Divide both sides by 3.}}\\ y^2 \amp = 25 \amp \amp \blert{\text{Take the square root of both sides.}}\\ y \amp = \pm \sqrt{25} \amp \amp \blert{\text{Simplify.}}\\ y \amp = \pm 5 \amp \amp \blert{\text{Remeber that a positive number has two square roots.}} \end{align*}
The solutions are \(5\) and \(-5\text{.}\)
🔗
🔗
🔗
🔗

Subsubsection A.3.1.3 Use geometric formulas

It may seem difficult to measure the inside of a round object like a sphere or a cone in cubic units, but you can imagine filling the object with liquid and then pouring the liquid into a box to measure its volume.
🔗
solids
Example A.3.9.
A cone is 16 cm tall and its volume is 500 cm\(^3\text{.}\) What is the radius of the base of the cone?
🔗
Solution.
We substitute \(h=16\) and \(V=500\) into the formula for the volume of a cone, and solve for \(r\text{.}\)
\begin{align*} \dfrac{1}{3} \pi r^2(\alert{16}) \amp = \alert{500} \amp \amp \blert{\text{Multiply both sides by 3.}}\\ 16 \pi r^2 \amp = 1500 \amp \amp \blert{\text{Divide both sides by}~16 \pi.}\\ r^2 \amp = 29.842 \amp \amp \blert{\text{Take the square root of both sides.}}\\ r \amp = 5.46 \end{align*}
The radius of the cone is 5.46 cm.
🔗
🔗
🔗
Checkpoint A.3.10.
A cylindrical syringe holds 100 cc (cubic centimeters) of fluid. If the syringe is 10 centimeters long, what is its radius?
🔗
Answer.
1.78 cm
🔗
🔗
🔗
Example A.3.11.
The diameter of a spherical wax candle is 5 inches. What is the volume of wax in the candle?
🔗
Solution.
The radius of the candle is half its diameter, or 2.5 inches. The volume of the candle is
\begin{equation*} V=\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi(2.5~ \text{in})^3=65.45~\text{in}^3 \end{equation*}
🔗
🔗
🔗
🔗

Subsubsection A.3.1.4 Use the Pythagorean theorem

If \(a\) and \(b\) are the lengths of the legs of a right triangle and \(c\) is the length of the hypotenuse, then
\begin{equation*} a^2 + b^2 = c^2 \end{equation*}
Note that the theorem is true only for right triangles -- ones that have a 90\(\degree\) angle.
🔗
Example A.3.14.
Find the unknown side in the right triangle.
🔗
right triangle
Solution.
The unknown side is the hypotenuse, so we apply the Pythagorean theorem with \(c=z,~a=12\text{,}\) and \(b=35\text{.}\)
\begin{align*} a^2 + b^2 \amp = c^2\\ 12^2+35^2 \amp =z^2 \amp \amp \blert{\text{Simplify the left side.}}\\ 1369 \amp = z^2 \amp \amp \blert{\text{Take the square root of both sides.}}\\ \pm 37 \amp = z \end{align*}
The length of the hypotenuse is a positive number, so \(z=37\text{.}\)
🔗
🔗
🔗
Example A.3.15.
A 25-foot ladder is placed against a wall so that its foot is 7 feet from the base of the wall. How far up the wall does the ladder reach?
🔗
Solution.
We make a sketch and label the known dimensions, calling the unknown height \(h\text{.}\) The ladder forms the hypotenuse of a right triangle, so we apply the Pythagorean theorem, substituting 25 for \(c\text{,}\) 7 for \(b\text{,}\) and \(h\) for \(a\text{.}\)
\begin{align*} a^2 + b^2 \amp = c^2\\ h^2 + 7^2 \amp = 25^2 \end{align*}
🔗
ladder
We solve the equation by extraction of roots:
\begin{align*} h^2 + 49 \amp = 625 \amp \amp \blert{\text{Subtract 49 from both sides.}}\\ h^2 \amp =576 \amp \amp \blert{\text{Extract roots.}}\\ h \amp =\pm \sqrt{576} \amp \amp \blert{\text{Simplify the radical.}}\\ h = \amp \pm 24 \end{align*}
The height must be a positive number, so the ladder reaches 24 feet up the wall.
🔗
🔗
🔗
🔗
🔗

Subsection A.3.2 Intercepts, Solutions, and Factors

In this lesson we review the skills we need to solve quadratic equations by factoring.
🔗

Subsubsection A.3.2.1 Multiply binomials

Example A.3.18.
Expand the product \(~(2x-3)(x-6)~\text{.}\)
🔗
Solution.
Multiply each term of the first binomial be each term of the second binomial. This gives four multiplications, often denoted by "FOIL," which stands for First terms, Outside terms, Inside terms, and Last terms.
🔗
\begin{align*} (2x+3)(x-6) \amp = \underline{2x \cdot x}~+~\underline{2x \cdot (-6)}~+~\underline{(-3) \cdot x}~+~\underline{(-3) \cdot (-6)}\\ \amp \quad \quad \blert{\text{F}}\quad \quad \quad \quad \blert{\text{O}} \quad \quad \quad \quad \quad \blert{\text{I}} \quad \quad \quad \quad \quad \blert{\text{L}}\\ \amp = 2x^2-12x-3x+18 ~~~~~~~~\quad\blert{\text{Combine like terms.}}\\ \amp = 2x^2-15x+18 \end{align*}
🔗
🔗
🔗
Example A.3.19.
Expand the product \(~-2(3x-4)(3x-5)\text{.}\)
🔗
Solution.
First, multiply the binomial factors together.
\begin{equation*} (3x-4)(3x-5) = 9x^2-27x+20 \end{equation*}
Then use the distributive law to multiply the result by the monomial factor, \(-2\text{.}\)
\begin{equation*} \alert{-2}(9x^2-27x+20) = -18x^2+54x-40 \end{equation*}
🔗
🔗
🔗
🔗

Subsubsection A.3.2.2 Factor quadratic trinomials

To factor the trinomial \(x^2+bx+c\text{,}\) we look for two numbers \(p\) and \(q\) whose product \(pq\) is the constant term and whose sum \(p+q\) is the coefficient of the middle term.
\begin{equation*} \begin{aligned}[t] (x+p)(x+q) \amp = x^2+qx+px+pq\\ \amp = x^2 +{\blert{(p+q)}}x +{\alert{pq}} = x^2+{\blert{b}}x+\alert{c} \end{aligned} \end{equation*}
🔗
Sign Patterns for Quadratic Trinomials.
Assume that \(b,~c,~p\) and \(q\) are positive integers. Then
  1. \(x^2+bx+c = (x+p)(x+q)\)
    🔗
    If all the coefficients of the trinomial are positive, then both \(p\) and \(q\) are positive.
    🔗
    🔗
  2. \(x^2-bx+c = (x-p)(x-q)\)
    🔗
    If the middle term of the trinomial is negative and the other two terms are positive, then \(p\) and \(q\) are both negative.
    🔗
    🔗
  3. \(x^2 \pm bx-c = (x+p)(x-q)\)
    🔗
    If the constant term of the trinomial is negative, then \(p\) and \(q\) have opposite signs.
    🔗
    🔗
🔗
🔗
Example A.3.24.
Factor \(~t^2+7t+12~\) as a product of two binomials,
\begin{equation*} t^2+7t+12 = (t+p)(t+q) \end{equation*}
🔗
Solution.
The constant term is \(12\text{,}\) so we look for two numbers \(p\) and \(q\) whose product is \(12\text{.}\) There are three possibilities:
\begin{equation*} 1~~ \text{and}~~ 12, ~~~~ 2 ~~\text{and} ~~6, ~~~~ \text{or} ~~~~ 3~~ \text{and}~~ 4 \end{equation*}
Because the middle term is \(7t\text{,}\) we must have \(p+q=7\text{.}\) We check each possibility and find that \(p=3\) and \(q=4\text{.}\) Thus,
\begin{equation*} t^2+7t+12 = (t+3)(t+4) \end{equation*}
🔗
🔗
🔗
Example A.3.25.
Factor \(~x^2-12x+20~\text{.}\)
🔗
Solution.
For this example we must find two numbers \(p\) and \(q\) for which \(pq=20\) and \(p+q=-12\text{.}\) These two conditions tell us that \(p\) and \(q\) must both be negative. We start by listing all the ways to factor \(20\) with negative factors:
\begin{equation*} -1~~\text{and}~~-20,~~~~-2~~\text{and}~~-10,~~~~-4~~\text{and}~~-5 \end{equation*}
We check \(p+q\) for each possibility to see which one gives the correct middle term. Because \(-2+(-10) = -12\text{,}\) the factorization is
\begin{equation*} x^2-12x+20 = (x-2)(x-10) \end{equation*}
🔗
🔗
🔗
Example A.3.26.
Factor \(~x^2+2x-15~\text{.}\)
🔗
Solution.
This time the product \(pq\) must be negative, so \(p\) and \(q\) must have opposite signs, one positive and one negative. There are only two ways to factor \(15\text{,}\) either \(1\) times \(15\) or \(3\) times \(5\text{.}\) We just "guess" that the second factor is negative, and check \(p+q\) for each possibility:
\begin{equation*} 1-15=-14~~~~~~\text{or}~~~~~~3-5=-2 \end{equation*}
The middle term we want is \(2x\text{,}\) not \(-2x\text{,}\) so we change the signs of \(p\) and \(q\text{:}\) we use \(-3\) and \(+5\text{.}\) The correct factorization is
\begin{equation*} x^2+2x-15 = (x-3)(x+5) \end{equation*}
🔗
🔗
🔗
🔗

Subsubsection A.3.2.3 Write algebraic expressions

Example A.3.33.
Ralph and Wanda together weigh 320 pounds. If Ralph weighs \(x\) pounds, how much does Wanda weigh?
🔗
Solution.
We subtract Ralph’s weight from the total; the remainder is Wanda’s weight: \(320-x\) pounds
🔗
🔗
🔗
Example A.3.34.
Delbert and Francine live 24 miles apart on Route 30. They meet at a cafe between their houses. If Delbert drove \(d\) miles, how far did Francine drive?
🔗
Solution.
We subtract Delbert’s distance from the total; the remainder is Francine’s distance: \(24-d\) miles
🔗
🔗
🔗
Example A.3.35.
Three eggs and two slices of buttered toast contain 526 calories. If one egg contains \(c\) calories, how many calories are in a slice of buttered toast?
🔗
Solution.
We subtract the calories in three eggs from the total; the remainder is the number of calories in two slices of toast, so one slice has half that many calories: \(~\frac{1}{2}(526-3c)\)
🔗
🔗
🔗
Example A.3.36.
The perimeter of a large rectangular playground is 124 yards. If its width is \(w\) yards, what is its length?
🔗
Solution.
We subtract twice the length from the perimeter; the remainder is twice the width, so the width is half that: \(~\frac{1}{2}(124-2w) = 62-w\) yards
🔗
🔗
🔗
Checkpoint A.3.38.
Six coffees and four pastries cost the office manager $21. If a pastry costs \(x\) dollars, how much does a coffee cost?
🔗
Answer.
\(\dfrac{1}{6}(21-4x)\) dollars
🔗
🔗
🔗
Checkpoint A.3.39.
The vertex angle of an isosceles triangle is \(v\) degrees. What is the measure of each of the two base angles?
🔗
Answer.
\(\dfrac{1}{2}(180-v)\) degrees
🔗
🔗
🔗
Checkpoint A.3.40.
The perimeter of a rectangular swimming pool is 260 feet. If the length of the pool is \(L\) feet, what is its width?
🔗
Answer.
\(\dfrac{1}{2}(260-2L)\) feet
🔗
🔗
🔗
🔗
🔗

Subsection A.3.3 Graphing Parabolas

Subsubsection A.3.3.1 Find the average of two numbers

The average of two numbers lies half-way between them on a number line. To find their average, we take one-half of their sum. That is, the average of \(p\) and \(q\) is
\begin{equation*} \dfrac{1}{2}(p+q)~~~~\text{or}~~~~\dfrac{p+q}{2} \end{equation*}
🔗
Example A.3.41.
The average of 4 and 9 is
\begin{equation*} \dfrac{1}{2}(4+9) = \dfrac{1}{2}(13) = \dfrac{13}{2},~~\text{or}~~6\dfrac{1}{2} \end{equation*}
🔗
🔗
Example A.3.42.
The average of \(-8\) and \(4\) is
\begin{equation*} \dfrac{1}{2}(-8+4) = \dfrac{1}{2}(-4) = -2 \end{equation*}
🔗
🔗
Example A.3.43.
The average of \(\dfrac{5}{2}\) and \(\dfrac{-3}{4}\) is
\begin{equation*} \dfrac{1}{2}\left(\dfrac{5}{2}-\dfrac{3}{4}\right) = \dfrac{1}{2}\left(\dfrac{10}{4}-\dfrac{3}{4}\right) = \dfrac{1}{2}\left(\dfrac{10}{4}\right) = \dfrac{7}{8} \end{equation*}
🔗
🔗
🔗

Subsubsection A.3.3.2 Solve quadratic equations

Example A.3.48.
Solve \(~3x^2=48\)
🔗
Solution.
We use extraction of roots. We first divide by 3 to isolate the squared expression.
\begin{align*} x^2 \amp = 16 \amp \amp \blert{\text{Take square roots.}}\\ x \amp = \pm 4 \end{align*}
The solutions are \(x=4\) and \(x=-4\text{.}\)
🔗
🔗
🔗
Example A.3.49.
Solve \(~3x^2=12x\)
🔗
Solution.
We solve by factoring. First, we get zero on one side of the equation.
\begin{align*} 3x^2-12x \amp = 0 \amp \amp \blert{\text{Factor the left side.}}\\ 3x(x-4) \amp = 0 \amp \amp \blert{\text{Set each factor equal to zero.}}\\ 3x=0,~~~~x-4 \amp = 0 \end{align*}
The solutions are \(x=0\) and \(x=4\text{.}\)
🔗
🔗
🔗
Example A.3.50.
Solve \(~3x^2-10x-8=0\)
🔗
Solution.
We solve by factoring. We factor the left side.
\begin{align*} (x-4)(3x+2) \amp = 0 \amp \amp \blert{\text{Set each factor equal to zero.}}\\ x-4=0,~~~~3x+2 \amp = 0 \amp \amp \blert{\text{Solve each equation.}}\\ x=4,~~~~x\amp = \dfrac{-2}{3} \end{align*}
The solutions are \(x=0\) and \(x=\dfrac{-2}{3}\text{.}\)
🔗
🔗
🔗
🔗

Subsubsection A.3.3.3 Find the coordinates of points on a parabola

To find the \(x\)-coordinate of a point on a parabola, we usually need to solve a quadratic equation.
🔗
Example A.3.55.
Find the \(y\)-coordinate of the point on the graph of \(~y=2x^2-3x+5~\) with \(x\)-coordinate \(-3\text{.}\)
🔗
Solution.
Substitute \(x=\alert{-3}\) into the equation, and evaluate.
\begin{equation*} y=2(\alert{-3})^2-3(\alert{-3})+5 = 18+9+5 = 32 \end{equation*}
The \(y\)-coordinate is 32, and the point is \((-3,32)\text{.}\)
🔗
🔗
🔗
Example A.3.56.
Find the \(x\)-coordinates of all points on the graph of \(~y=20-3x^2~\) with \(y\)-coordinate \(-28\text{.}\)
🔗
Solution.
Substitute \(y=\alert{-28}\) into the equation, and solve.
\begin{align*} \alert{-28} \amp = 20-3x^2 \amp \amp \blert{\text{Subtract 20 from both sides.}}\\ -48 \amp =-3x^2 \amp \amp \blert{\text{Divide both sides by}~-3.}\\ 16 \amp = x^2 \amp \amp \blert{\text{Extract roots.}}\\ \pm 4 \amp = x \end{align*}
The points are \((4, -28)\) and \((-4,-28)\text{.}\)
🔗
🔗
🔗
Checkpoint A.3.58.
The \(x\)-coordinate of the vertex of \(~y=2x^2-6x+1~\) is \(\dfrac{3}{2}\text{.}\) Find the \(y\)-coordinate of the vertex.
🔗
Answer.
\(\dfrac{-7}{2}\)
🔗
🔗
🔗
🔗
🔗

Subsection A.3.4 Completing the Square

Subsubsection A.3.4.1 Squares of binomials

To solve a quadratic equation by completing the square, we create the square of a binomial:
\begin{equation*} \blert{(x+p)^2 = x^2 + 2px + p^2} \end{equation*}
🔗
Example A.3.61.
Write each trinomial as the square of a binomial.
  1. \(\displaystyle x^2+6x+9\)
    🔗
  2. \(\displaystyle x^2-5x+\dfrac{25}{4}\)
    🔗
🔗
Solution.
  1. In the formula above, note that the coefficient of \(x\) is \(2p\) and the constant term is \(p^2\text{.}\) In this example, \(2p=6\) and \(p^2=9\text{,}\) so \(p=\alert{3}\text{,}\) and
    \begin{equation*} x^2+6x+9 = (x+\alert{3})^2 \end{equation*}
    🔗
    🔗
  2. The coefficient of \(x\) is \(2p=-5\) and the constant terms is \(p^2=\dfrac{25}{4}\text{,}\) so \(p=\alert{\dfrac{-5}{2}}\text{,}\) and
    \begin{equation*} x^2-5x+\dfrac{25}{4} = \left(x-\alert{\dfrac{5}{2}}\right)^2 \end{equation*}
    🔗
    🔗
🔗
🔗
🔗
🔗
To solve a quadratic equation by completing the square, we often have to work with fractions.
🔗

Subsubsection A.3.4.2 Multiply fractions

To multiply two fractions together, we multiply their numerators together, and multiply their denominators together. We can divide out any common factors in numerator and denominator before we multiply.
🔗
Example A.3.66.
Multiply \(~\dfrac{3}{8} \cdot \dfrac{6}{5}\)
🔗
Solution.
We can divide out a factor of 2.
\begin{equation*} \dfrac{3}{8} \cdot \dfrac{6}{5} = \dfrac{3}{\cancel{2} \cdot 4} \cdot \dfrac{\cancel{2} \cdot 3}{5} = \dfrac{9}{20} \end{equation*}
🔗
🔗
🔗
Example A.3.67.
Multiply \(~\dfrac{ab}{6} \cdot \dfrac{3a}{2b}\)
🔗
Solution.
We divide out common factors before multiplying.
\begin{equation*} \dfrac{ab}{6} \cdot \dfrac{3a}{2b} = \dfrac{a \cancel{b}}{2 \cdot \cancel{3}} \cdot \dfrac{\cancel{3}a}{2 \cancel{b}} = \dfrac{a^2}{4} \end{equation*}
🔗
🔗
🔗
🔗

Subsubsection A.3.4.3 Add fractions

To add or subtract unlike fractions.
  1. Find the LCD for the fractions.
    🔗
  2. Build each fraction to an equivalent one with the LCD as its denominator.
    🔗
  3. Add or subtract the numerators. Keep the same denominator.
    🔗
🔗
🔗
Example A.3.70.
Add.
  1. \(\displaystyle \dfrac{7}{10} + \dfrac{5}{6}\)
    🔗
  2. \(\displaystyle 6 + \dfrac{4}{9}\)
    🔗
🔗
Solution.
  1. Step1: Find the LCD. Factor each denominator.
    \begin{gather*} 10=2\cdot 5\\ 6 = 2\cdot 3 \end{gather*}
    The LCD is \(~2 \cdot 3 \cdot 5 = 30\text{.}\)
    🔗
    Step2: Build each fraction to a denominator of 30. The building factor for the first fraction is \(\blert{3}\text{,}\) and \(\blert{5}\) for the second fraction.
    \begin{equation*} \dfrac{7}{10} \cdot \blert{\dfrac{3}{3}} = \dfrac{21}{30}~~~~\text{and}~~~~\dfrac{5}{6} \cdot \blert{\dfrac{5}{5}} = \dfrac{25}{30} \end{equation*}
    Step 3: Add the two like fractions, and reduce.
    \begin{align*} \dfrac{7}{10} + \dfrac{5}{6} \amp = \dfrac{21}{30} + \dfrac{25}{30} = \dfrac{46}{30}\\ \amp = \dfrac{\cancel{2} \cdot 23}{\cancel{2} \cdot 15} = \dfrac{23}{15} \end{align*}
    🔗
    🔗
  2. Step1: The LCD is 9.
    🔗
    Step 2: Build the whole number to a denominator of 9.
    \begin{equation*} \dfrac{6}{1} \cdot \blert{\dfrac{9}{9}} = \dfrac{54}{9} \end{equation*}
    Step 3: Add the two like fractions.
    \begin{equation*} 6 + \dfrac{4}{9} = \dfrac{54}{9} + \dfrac{4}{9} = \dfrac{58}{9} \end{equation*}
    🔗
    🔗
🔗
🔗
🔗
🔗
🔗
You have attempted of activities on this page.
🔗
PrevTopNext