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lukranit [14]
3 years ago
8

A three-dimensional figure was constructed using identical cubes. The top-, front-, and left-side views of this figure are shown

. Which could be this figure?

Mathematics
2 answers:
Strike441 [17]3 years ago
5 0

Answer:

The one above the one you selected

Step-by-step explanation:

krek1111 [17]3 years ago
3 0

Answer:

たの愚か者私はあなたを理解していません

Step-by-step explanation:

You might be interested in
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
Aaron's car used 3- gallons of gas when
monitta
We want to find miles per gallon. Another way of saying miles per gallon is miles/gallon. So we will plug the numbers into this equation:

60 miles/3 gallons = 20 miles/gallon

This is because 60/3=20.

I hope this makes sense.
4 0
3 years ago
Read 2 more answers
Area A, equals the product of b and h divided by 2, slove for h
gizmo_the_mogwai [7]
A = bh/2
2a = bh
2a/b = h

h = 2a/b
7 0
3 years ago
What is the volume of a cube with a length of 4/5 m?
a_sh-v [17]

Answer:

64/125 cubic meter

Step-by-step explanation:

volume = l × b × h

volume of a cube = s × s × s

∴ 4/5 × 4/5 × 4/5 = 64/125

8 0
3 years ago
Please help me out :P
Veseljchak [2.6K]

cos θ = \frac{-4\sqrt{65} }{65}, sin θ = \frac{-7\sqrt{65} }{65}, cot  θ  = 4/7, sec  θ = \frac{-\sqrt{65} }{4}, cosec  θ  = \frac{-\sqrt{65} }{7}

<h3>What are trigonometric ratios?</h3>

Trigonometric Ratios are values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin θ: Opposite Side to θ/Hypotenuse

Tan θ: Opposite Side/Adjacent Side & Sin θ/Cos

Cos θ: Adjacent Side to θ/Hypotenuse

Sec θ: Hypotenuse/Adjacent Side & 1/cos θ

Analysis:

tan θ = opposite/adjacent = 7/4

opposite = 7, adjacent = 4.

we now look for the hypotenuse of the right angled triangle

hypotenuse = \sqrt{7^{2} + 4^{2} } = \sqrt{49+16} = \sqrt{65}

sin θ = opposite/ hyp = \frac{7}{\sqrt{65} }

Rationalize, \frac{7}{\sqrt{65} } x \frac{\sqrt{65} }{\sqrt{65} } = \frac{7\sqrt{65} }{65}

But θ is in the third quadrant(180 - 270) and in the third quadrant only tan and cot are positive others are negative.

Therefore, sin θ = - \frac{7\sqrt{65} }{65}

cos   θ  = adj/hyp = \frac{4}{\sqrt{65} }

By rationalizing and knowing that cos  θ  is negative, cos θ  = -\frac{-4\sqrt{65} }{65}

cot θ  = 1/tan θ  = 1/7/4 = 4/7

sec θ  = 1/cos θ  = 1/\frac{4}{\sqrt{65} } = -\frac{-\sqrt{65} }{4}

cosec θ  = 1/sin θ  = 1/\frac{\sqrt{65} }{7} = \frac{-\sqrt{65} }{7}

Learn more about trigonometric ratios: brainly.com/question/24349828

#SPJ1

5 0
1 year ago
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