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exis [7]
3 years ago
6

Find the total surface area of this square based pyramid. 10 in 5 in

Mathematics
2 answers:
Nana76 [90]3 years ago
7 0

Answer:

128.08 in^2

Step-by-step explanation:

To find the surface area use the expression a^2 + 2a\sqrt{\frac{a^2}{4} +h^2}

When you plug in 5 for a and 10 for h you will get 128.08

zzz [600]3 years ago
4 0

Answer:

  125 in²

Step-by-step explanation:

Each triangular face has a base of 5 in and a height of 10 in. The area of it is given by the formula ...

  A = (1/2)bh

  A = (1/2)(5 in)(10 in) = 25 in²

The square base has an area given by the formula ...

  A = s² . . . . . where s is the side length

  A = (5 in)² = 25 in²

The total area is the sum of the areas of the 4 faces and the base:

  total area = 4 × (area of 1 face) + (area of base)

  total area = 4 × (25 in²) + 25 in²

  total area = 125 in²

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Problem 35

The bar graph is shown below

You simply draw various rectangles such that the heights represent the frequency of each animal type.

Eg: there are 20 elephants, so the "elephant" bar is 20 units tall.

You can make the bar graph by hand, or use spreadsheet software. I recommend going with software (if you can).

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

Problem 36

1 book = 20 mm thickness

5 books = 5*20 = 100 mm thickness

1 paper = 0.016 mm thickness

5 papers = 5*0.016 = 0.08 mm thickness

total thickness = 100+0.08 = 100.08 mm

<h3>Answer: 100.08 mm</h3>

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

Problem 37

121/11 = 11

121 = 11*11

If we say 11+11+11+...+11, and have 11 copies of these values added, then we get to a sum of 121

This is probably the easiest way to get the answer assuming repeated values are allowed.

You can stop here if your teacher allows you to use repeated values. If not, then move onto the next section.

-----------

If your teacher requires you to add 11 <u>different</u> numbers, then you can follow this procedure

  1. Write out eleven copies of "11" in a sequence
  2. Subtract 2 from the first "11" (to get 9) and add it to the last copy of "11" (to get 13)
  3. Subtract 4 from the second "11" (to get 7) and add it to the second to last copy of "11" (to get 15)
  4. Subtract 6 from the third "11" (to get 5) and add it to the third to last copy of "11" (to get 17)
  5. Subtract 8 from the fourth "11" (to get 3) and add it to the fourth to last copy of "11" (to get 19)
  6. Finally, subtract 10 from the fifth "11" (to get 1) and add it to the fifth to last copy of "11" (to get 21)

After carefully following those steps, you'll get this sequence (in the exact order shown):

{9, 7, 5, 3, 1, 11, 21, 19, 17, 15, 13}

There are three properties to notice of this sequence

  1. It's composed of two decreasing arithmetic sequences {9, 7, 5, 3, 1} and {21, 19, 17, 15, 13}
  2. The 11 in the middle hasn't been changed from the original sequence of nothing but "11"s.
  3. You should find that the terms of this new sequence add to 121.

That sequence we got sorts to {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21}

and we can say 1+3+5+7+9+11+13+15+17+19+21 = 121

<h3>Answer: 1+3+5+7+9+11+13+15+17+19+21 = 121</h3>

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3 years ago
The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit at which the planet is cl
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Consider the attached ellipse. Let the sun be at the right focus. Then perihelion is at right vertex on the x-axis and aphelion is at the left vertex on the x-axis.

The distances:

  • from perihelion to the sun in terms of ellipse is a-c;
  • from aphelion to the sun in terms of ellipse is a+c.

Then

\left\{\begin{array}{l}a-c=741,000,000\\a+c=817,000,000\end{array}\right.

Add these two equations:

2a=1,558,000,000 \\ \\a=779,000,000

and subtract first equation from the second:

2c=76,000,000 \\ \\c=38,000,000.

Note that b=\sqrt{a^2-c^2}, thus

b=\sqrt{779,000,000^2-38,000,000^2}=\sqrt{605,397,000,000,000,000}.

The equation for the planet's orbit is

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\Rightarrow \dfrac{x^2}{606,841,000,000,000,000}+\dfrac{y^2}{605,397,000,000,000,000}=1.

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