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3 years ago

twelve rolls of commercial-grade paper towels come in one case costing $23.58 What is the the cost per roll of paper towels

Mathematics

2 answers:

Aloiza [94]3 years ago

7 0

You would take your cost of 23.58 and divide it by the amount (12). You would end up getting 1.965 or 1.97 per roll

soldi70 [24.7K]3 years ago

3 0

$23.58 divided by 12 = $1.965
Each roll is about $1.97 (if you round up)

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3 years ago

The length of a rectangle is the width minus 8 units. The area of the rectangle is 9 units. What is the length, in units, of the

frez [133]

Length of rectangle is 1 unit

<u>Solution: </u>

Given that

Length of a rectangle is width minus 8 units.

Area of rectangle = 9 square units

Need to calculate the length of rectangle.

Let us assume width of rectangle = x

As Length is width minus 8,

Length of the rectangle = x – 8

$\text{ Area of rectangle }=\text{ width of rectangle }\times \text{ Length of the rectangle }$

$\text{ Area of rectangle }= x\times (x-8)$

$\text { Area of rectangle }=\left(x^{2}-8 x\right)$

As given that area of rectangle = 9 square units

$\begin{array}{l}{\Rightarrow x^{2}-8 x=9} \\\\ {\Rightarrow x^{2}-8 x-9=0}\end{array}$

On solving above quadratic equation for x, we get

$\begin{array}{l}{\Rightarrow x^{2}-9 x+x-9=0} \\\\ {\Rightarrow x(x-9)+1(x-9)=0} \\\\ {\Rightarrow (x-9)(x+1)=0}\end{array}$

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3 years ago

1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat

brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

$x = 2$ is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

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The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

5 0

2 years ago