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timama [110]
1 year ago
10

The Pyramid of Khufu in Giza, Egypt was the world's tallest free-standing structure for more than 3,500 years. Its original heig

ht was about 144
meters. Its base is approximately a square with a side length of 231 meters.
The diagram shows a cross section created by dilating the base using the top of the pyramid as the center of dilation. The cross section is at a
height of 96 meters
144 m
=> A
96 m
231 m
231 m
Complete the following about the cross section
The scale factor used to create the cross section is 13
Part B
Complete the following about the cross section
The dimensions of the cross section is a Select Choice y meter by Select Choice y meter Select Choice

Mathematics
1 answer:
Hoochie [10]1 year ago
7 0

The scale factor used to create the cross section is 1/3. The cross section has a base side length of 77 meters.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

The height of the dilated pyramid = 144 m - 96 m = 48 m

Scale factor of dilation = 48 m / 144 m = 1/3

The side length of dilated pyramid = (1/3) * 231 m = 77 m

The scale factor used to create the cross section is 1/3. The cross section has a base side length of 77 meters.

Find out more on equation at: brainly.com/question/2972832

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Step-by-step explanation:

x+7=19

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Need help !!!Divide 5a^2 + 6a - 9 into 25a^4 + 19a^2 - a - 78.
lana [24]
Hello,

Maybe i have not understood "into".

\dfrac{25a^4+19a^2-a-78}{5a^2+6a-9} =5a^2-6a+20+ \dfrac{-175a+102}{5a^2+6a-9}
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WHO CAN solve it Please !
Mariulka [41]

Answer:

a) True

<em>      </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha =0<em></em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the definite integration</em>

<em>            </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha<em></em>

<em>we know that the trigonometric formula</em>

<em> sin²∝+cos²∝ = 1</em>

<em>            cos²∝ = 1-sin²∝</em>

<u><em>step(ii):-</em></u>

<em>Now the  integration</em>

<em>         </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^{2} \alpha } } \, )d\alpha<em></em>

<em>                                      = </em>\int\limits^\pi _0 {cos\alpha } \, dx<em></em>

<em>Now, Integrating </em>

<em>                                  </em>= ( sin\alpha )_{0} ^{\pi }<em></em>

<em>                                = sin π - sin 0</em>

<em>                               = 0-0</em>

<em>                              = 0</em>

<u><em>Final answer:-</em></u>

<em>      </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha =0<em></em>

<em></em>

<em></em>

5 0
2 years ago
The sum of three consecutive even numbers is one hundred sixty-two. What is the smallest of the three numbers?
zepelin [54]
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x + y + z = 168

(y-2) + y + (y+2) = 168.

3y = 168

y = 56

therefore the smallest number (x) = 54.
436 viewsView upvotes

1




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