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Flauer [41]
3 years ago
11

2 more than 5 groups of 3

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
5 0
5*3+2 = 15+2 = 17
The asnwer is 17.
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All boxes with a square​ base, an open​ top, and a volume of 60 ft cubed have a surface area given by ​S(x)equalsx squared plus
Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval (0,\infty) is S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2

The dimensions of the box with minimum surface​ area are: the base edge x=2\sqrt[3]{15}\:ft and the height h=\sqrt[3]{15} \:ft

Step-by-step explanation:

We are given the surface area of a box S(x)=x^2+\frac{240}{x} where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval (0,\infty) and the dimensions of the box with minimum surface​ area.

1. To find the absolute minimum you must find the derivative of the surface area (S'(x)) and find the critical points of the derivative (S'(x)=0).

\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}

Next,

2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120

There is a undefined solution x=0 and a real solution x=2\sqrt[3]{15}. These point divide the number line into two intervals (0,2\sqrt[3]{15}) and (2\sqrt[3]{15}, \infty)

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before x=2\sqrt[3]{15}, increases after it, and is defined at x=2\sqrt[3]{15}. So f(x) has a relative minimum point at x=2\sqrt[3]{15}.

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for x=2\sqrt[3]{15}.

\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}

and for x=2\sqrt[3]{15} we get:

2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0

Therefore S(x) has a minimum at x=2\sqrt[3]{15} which is:

S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2

2. To find the third dimension of the box with minimum surface​ area:

We know that the volume is 60 ft^3 and the volume of a box with a square base is V=x^2h, we solve for h

h=\frac{V}{x^2}

Substituting V = 60 ft^3 and x=2\sqrt[3]{15}

h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft

The dimension are the base edge x=2\sqrt[3]{15}\:ft and the height h=\sqrt[3]{15} \:ft

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What is the equation of the line that has a slope of 3 and a​ y-intercept of​ 2?
svetoff [14.1K]

The equation of a line is written as t = ax + b where a is the slope and b is the y-intercept.

Using the slope and y intercept given the equation is:

Y = 3x + 2

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Answer:

Inequality Form: x > -17

Interval Notation: (-17, ∞)

Step-by-step explanation:

brainliest would be great

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??? what steps are you talking about. lol

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