Question

15 the volume v(x) of a box is defined by the function v(x)=x(16 - x)(18 - 4x), where each factor represents a dimension of the box. what is the domain of the function for this problem situation? a. 0<x<4 b. 0<x<4.5 c. 4<x<16 d. 4.5<x<16 16 the product of two linear - function factors produces a quadratic function that is concave up. which could be the set of linear - function factors? select all that apply. a. f(x)=-2x - 3 and g(x)=x - 2 b. f(x)=x - 1/3 and g(x)=1/2x + 4 c. f(x)=-26 and g(x)=-3x + 1 d. f(x)=-1/5x + 3 and g(x)=-4x - 5 17 which functions have a zero with a multiplicity of 3? select all that apply. a. (x - 2)^2 b. (x - 2)^3(x - 3)(x + 4) c. (x^2 + 1)(x + 5)(2x - 6) d. (x - 1)(x - 1)(x + 1) 18 the function v(h) models the volume of a planter box with height h. at which height does the planter box have a volume of 50 cubic inches? select all that apply. a. 0.25 in. b. 4.76 in. c. 5.36 in. d. 9.39 in.

Explanation:

15.

Step1: Consider non - negative dimensions

Since the dimensions of a box must be non - negative, we set each factor in \(V(x)=x(16 - x)(18 - 4x)\) greater than or equal to 0.

1. \(x\geq0\).
2. \(16 - x\geq0\), which gives \(x\leq16\).
3. \(18 - 4x\geq0\), solving for \(x\) we get \(4x\leq18\) or \(x\leq4.5\). Also, since \(x\) represents a dimension, \(x>0\).

The product of two linear functions \(f(x)=ax + b\) and \(g(x)=cx + d\) gives a quadratic function \(y=(ax + b)(cx + d)=acx^{2}+(ad + bc)x+bd\). A quadratic function \(y = Ax^{2}+Bx + C\) is concave up when \(A>0\).

Step1: Check option a

For \(f(x)=-2x - 3\) and \(g(x)=x - 2\), the product is \((-2x - 3)(x - 2)=-2x^{2}+4x-3x + 6=-2x^{2}+x + 6\), \(A=-2<0\), so it's not concave up.

Step2: Check option b

For \(f(x)=x-\frac{1}{3}\) and \(g(x)=\frac{1}{2}x + 4\), the product is \((x-\frac{1}{3})(\frac{1}{2}x + 4)=\frac{1}{2}x^{2}+4x-\frac{1}{6}x-\frac{4}{3}=\frac{1}{2}x^{2}+\frac{23}{6}x-\frac{4}{3}\), \(A = \frac{1}{2}>0\), so it's concave up.

Step3: Check option c

\(f(x)=-26\) is a constant function, not a linear function, so this option is incorrect.

Step4: Check option d

For \(f(x)=-\frac{1}{5}x + 3\) and \(g(x)=-4x - 5\), the product is \((-\frac{1}{5}x + 3)(-4x - 5)=\frac{4}{5}x^{2}+x-12x - 15=\frac{4}{5}x^{2}-11x - 15\), \(A=\frac{4}{5}>0\), so it's concave up.

The zero of a function \(y=(x - r)^n\) has multiplicity \(n\).

Step1: Analyze option a

For \(y=(x - 2)^{2}\), the zero \(x = 2\) has multiplicity 2.

Step2: Analyze option b

For \(y=(x - 2)^{3}(x - 3)(x + 4)\), the zero \(x = 2\) has multiplicity 3.

Step3: Analyze option c

For \(y=(x^{2}+1)(x + 5)(2x - 6)\), \(x^{2}+1\) has no real zeros, \(x=-5\) has multiplicity 1 and \(x = 3\) has multiplicity 1.

Step4: Analyze option d

For \(y=(x - 1)(x - 1)(x + 1)=(x - 1)^{2}(x + 1)\), the zero \(x = 1\) has multiplicity 2.

Answer:

b. \(0 < x<4.5\)

16.