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DochEvi [55]
3 years ago
8

A square pyramid has a base with an area of 20 square meters, and its lateral faces have a slant height of x meters. Sydney is c

onstructing a second square pyramid with the same size base, but the lateral faces of her pyramid have a slant height twice as long, 2x. Which statement best describes how the surface area of Sydney’s pyramid compares to the surface area of the original pyramid?

Mathematics
2 answers:
Lady bird [3.3K]3 years ago
6 0

Answer:

The answer is D, "Sydney's pyramid will have a surface area that is greater than the original pyramid's but not double the area because the slant height is not used when finding the area of the base." (Just took the test)

Step-by-step explanation:


svetoff [14.1K]3 years ago
5 0
1. Consider the pyramid ABCDE in the figure 1 attached:

Each side of the square base equals \sqrt{20} = \sqrt{4*5} = \sqrt{4}* \sqrt{5}=2 \sqrt{5} (units)

Area of 1 lateral side of pyramid ABCDE is \frac{1}{2}* 2 \sqrt{5}*x=\sqrt{5}x

so the total lateral area is 4\sqrt{5}x (units squared)

and the total surface area is 20+4\sqrt{5}x (units squared)

2. Now consider the second figure KLMNP, Sydney's pyramid:

The lateral area is 4* \frac{1}{2}* 4 \sqrt{5} *2x=16 \sqrt{5}x

so the total area of the second pyramid is 20+16 \sqrt{5}x

So 2 things can be said about the areas:

i) the lateral area of Sydney's pyramid is \frac{16 \sqrt{5}x}{4\sqrt{5}x} =4 times larger than the lateral area of the original pyramid.

ii) The total area of Sydney's pyramid is (20+16 \sqrt{5}x)-(20+4\sqrt{5}x)=12\sqrt{5}x units squared larger than the total area of the original pyramid.

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Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.
igor_vitrenko [27]

Answer:

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7

So, the solution of the equation is

y = Ae^{m_{1} t} + Be^{m_{2} t}

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So,

y = Ae^{3t} + Be^{-7t}

Also,

y' = 3Ae^{3t} - 7e^{-7t}

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1      (1)

y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}

Substituting A into (1) above, we have

\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1      \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}

Substituting B into A, we have

A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}

Substituting A and B into y, we have

y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

So the solution to the differential equation is

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

6 0
3 years ago
What’s the answer to this question?
GarryVolchara [31]
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3 is x
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5 = 2 x 3 - 1

5= 6 -1

Which is 5
7 0
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