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3 years ago

What is the surface area and volume? PLEASE EXPLAIN IN DETAIL HOW TO SOLVE.

Mathematics

1 answer:

Maslowich3 years ago

7 0

The surface area is the sum of the area of all the faces of an object.

Area of bottom face: 4×2.5=10

Area of right and left sides: 2(2.5×2)=10

Area of front and back sides: 2(4×2)=16

Area of the rectangular sides of the roof: 2(2.5×2.5)=12.5

Area of the triangle sides of the roof: 2(1/2(4×1.5))=6

Add up all of these to get 54.5 yd^2

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How do you find all the solutions to the equation 5x^2-27=2x^2+48 using the zero product property?

Dmitrij [34]

First, you have to get the equation equal to 0. To do that, you need to subtract the $2 x^{2}$ and 48 from the right side of the equation and put it on the left side. The new equation will read

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3 years ago

Read 2 more answers

What is the coefficient in the expression x-27?

Ymorist [56]

X - 27

The coefficient is a number used to multiply a variable. The variable is x. The coefficient is : (understood but not written)...it is 1.
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3 years ago

Mrs. Renz needed to by mulch and soil for the prayer garden and was charged by the pound. She bought five bags of mulch and thre

dolphi86 [110]

Answer:

3.25 pounds

Step-by-step explanation:

First subtract the pounds of soil 2.2*3=6.6 so that takes away 6.6

22.85-6.6 =16.25

now divide 16.25 by 5

you get 3.25 so each bag was 3.25 pounds

and you multiply/add just to check

6 0

4 years ago

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 $h=\sqrt[3]{15} \:ft$

6 0

3 years ago

Margaret is planning a wedding reception. She wants to spend less than her budget of $5,748, and Margaret has already spent $2,4

noname [10]

Answer:

She can invite 109 guests and be right on top of budget. If she wants to spend less than her budget though she can only invite 108 people.

Step-by-step explanation:

First you have to subtract 5,748 - 2,478 which is 3270. Next you have to divide 3270 by 30 because each person costs 30 dollars. 3270 divided by 30 is 109.

I hope this helps! Please mark me brainliest if I am correct! Have a nice day!

6 0

3 years ago