Question

the graph of (y = f(x)) is shown, where the function (f) is defined by (f(x)=ax^{3}+bx^{2}+cx + d) and (a,b,c,) and (d) are constants. for how many values of (x) does (f(x)=0)?
a) one
b) two
c) three
d) four

the given equation (p=\frac{k}{4j + 9}) relates the distinct positive numbers (p,k,) and (j). which equation correctly expresses (4j + 9) in terms of (p) and (k)?
a) (4j + 9=\frac{k}{p})
b) (4j + 9=kp)
c) (4j + 9=k - p)
d) (4j + 9=\frac{p}{k})

the area (a), in square - centimeters, of a rectangular cutting - board can be represented by the expression (w(w + 9)), where (w) is the width, in centimeters, of the cutting - board. which expression represents the length, in centimeters, of the cutting - board?
a) (w(w + 9))
b) (w)
c) (9)
d) ((w + 9))

circle (a) has a radius of (3n) and circle (b) has a radius of (129n), where (n) is a positive constant. the area of circle (b) is how many times the area of circle (a)?
a) 43
b) 86
c) 129
d) 1,849

Explanation:

1. First question (graph - related):

• Explanation:
• The solutions of the equation \(f(x)=0\) are the \(x\) - intercepts of the graph of \(y = f(x)\). By looking at the graph of \(y=f(x)\), we count the number of times the graph crosses the \(x\) - axis.
• The graph of \(y = f(x)\) crosses the \(x\) - axis at three distinct points.
• Answer: C. Three

1. Second question (algebraic - equation manipulation):

• Explanation:
• Given the equation \(p=\frac{k}{4j + 9}\).
• Step1: Cross - multiply
• Multiply both sides of the equation by \(4j + 9\) to get \(p(4j + 9)=k\).
• Step2: Solve for \(4j + 9\)
• Divide both sides of the equation \(p(4j + 9)=k\) by \(p\) (since \(p

eq0\) as it is a non - zero positive number), we have \(4j + 9=\frac{k}{p}\).

• Answer: A. \(4j + 9=\frac{k}{p}\)

1. Third question (area of rectangle - algebraic):

• Explanation:
• The area \(A\) of a rectangle is given by \(A=\text{length}\times\text{width}\). Here, \(A = w(w + 9)\) and the width is \(w\).
• Let the length be \(L\). Then \(A = L\times w\), and since \(A=w(w + 9)\), dividing both sides by \(w\) (assuming \(w

eq0\)), we get \(L=w + 9\).

• Answer: D. \((w + 9)\)

1. Fourth question (area of circles - geometric and algebraic):

• Explanation:
• The area formula of a circle is \(A=\pi r^{2}\).
• The area of circle \(A\) with radius \(r_A = 3n\) is \(A_A=\pi(3n)^{2}=9\pi n^{2}\).
• The area of circle \(B\) with radius \(r_B = 129n\) is \(A_B=\pi(129n)^{2}=16641\pi n^{2}\).
• To find how many times the area of circle \(B\) is the area of circle \(A\), we calculate \(\frac{A_B}{A_A}=\frac{16641\pi n^{2}}{9\pi n^{2}}\).
• Step1: Cancel out common terms
• Cancel out \(\pi\) and \(n^{2}\) in the numerator and denominator. We get \(\frac{16641}{9}=1849\).
• Answer: D. 1849

Answer:

1. First question (graph - related):

• Explanation:
• The solutions of the equation \(f(x)=0\) are the \(x\) - intercepts of the graph of \(y = f(x)\). By looking at the graph of \(y=f(x)\), we count the number of times the graph crosses the \(x\) - axis.
• The graph of \(y = f(x)\) crosses the \(x\) - axis at three distinct points.
• Answer: C. Three

1. Second question (algebraic - equation manipulation):

• Explanation:
• Given the equation \(p=\frac{k}{4j + 9}\).
• Step1: Cross - multiply
• Multiply both sides of the equation by \(4j + 9\) to get \(p(4j + 9)=k\).
• Step2: Solve for \(4j + 9\)
• Divide both sides of the equation \(p(4j + 9)=k\) by \(p\) (since \(p

eq0\) as it is a non - zero positive number), we have \(4j + 9=\frac{k}{p}\).

• Answer: A. \(4j + 9=\frac{k}{p}\)

1. Third question (area of rectangle - algebraic):

• Explanation:
• The area \(A\) of a rectangle is given by \(A=\text{length}\times\text{width}\). Here, \(A = w(w + 9)\) and the width is \(w\).
• Let the length be \(L\). Then \(A = L\times w\), and since \(A=w(w + 9)\), dividing both sides by \(w\) (assuming \(w

eq0\)), we get \(L=w + 9\).

• Answer: D. \((w + 9)\)

1. Fourth question (area of circles - geometric and algebraic):

• Explanation:
• The area formula of a circle is \(A=\pi r^{2}\).
• The area of circle \(A\) with radius \(r_A = 3n\) is \(A_A=\pi(3n)^{2}=9\pi n^{2}\).
• The area of circle \(B\) with radius \(r_B = 129n\) is \(A_B=\pi(129n)^{2}=16641\pi n^{2}\).
• To find how many times the area of circle \(B\) is the area of circle \(A\), we calculate \(\frac{A_B}{A_A}=\frac{16641\pi n^{2}}{9\pi n^{2}}\).
• Step1: Cancel out common terms
• Cancel out \(\pi\) and \(n^{2}\) in the numerator and denominator. We get \(\frac{16641}{9}=1849\).
• Answer: D. 1849