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Galina-37 [17]
3 years ago
9

At a height of about 212 meters, One Shell Square is the tallest building in New Orleans. Marlie is creating a scale model of th

is building, using the scale 250 meters : 1 meter.
To the nearest tenth of a meter, what will be the height of the scale model?

Enter your answer as a decimal in the box.
Mathematics
2 answers:
KengaRu [80]3 years ago
8 0

Answer:

0.8

Step-by-step explanation:

212 / 250 = 0.848

0.848 x 1 = 0.848

The answer has to be to the nearest tenth of a meter so we have to cut of the 0.048 off of 0.848 giving us 0.8

<em>x</em> = 0.8

Please give me brainliest

sveticcg [70]3 years ago
6 0
For the answer to the question above asking to the nearest tenth of a meter, what will be the height of the scale model if a<span>At a height of about 212 meters, One Shell Square is the tallest building in New Orleans. Marlie is creating a scale model of this building, using the scale 250 meters : 1 meter.
My answer would be 0.8m</span>
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What is the y-intercept of y = 5x - 1?​
Serhud [2]

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Given Equation: y=5x-1

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8 0
4 years ago
Hello people ~
Luden [163]

Cone details:

  • height: h cm
  • radius: r cm

Sphere details:

  • radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10²                                   [insert values]

r² = 10² - (h-10)²                                     [change sides]

r² = 100 - (h² -20h + 100)                       [expand]

r² = 100 - h² + 20h -100                        [simplify]

r² = 20h - h²                                          [shown]

r = √20h - h²                                       ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (\sqrt{20h - h^2})^2  \  ( h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20h - h^2)  (h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20 - h) (h) ( h)

\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )

<u>apply chain rule</u>

\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0

\Longrightarrow \sf 40h-3h^2=0

\Longrightarrow \sf h(40-3h)=0

\Longrightarrow \sf h=0, \ 40-3h=0

\Longrightarrow \sf  h=0,\:h=\dfrac{40}{3}<u />

<u>maximum volume:</u>                <u>when h = 40/3</u>

\sf \Longrightarrow max=  \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )

\sf \Longrightarrow maximum= 1241.123 \ cm^3

<u>minimum volume:</u>                 <u>when h = 0</u>

\sf \Longrightarrow min=  \dfrac{1}{3} \pi (0)^2(20-0)

\sf \Longrightarrow minimum=0 \ cm^3

6 0
2 years ago
Read 2 more answers
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