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eimsori [14]
3 years ago
15

Pyramids and cones (finding surface area). Please help!

Mathematics
1 answer:
zimovet [89]3 years ago
7 0
For the first one

notice, the pyramid is just 4 triangles and one square, stacked up to each other at the edges

now, for the first one, you have a square 4x4 and four triangles or base 4 and height of 6

bear in mind, the pyramids are called "regular pyramids" because their base is a regular polygon, meaning, a polygon with all equal sides, in this case a quadrilteral, or a square

---------------------------------------------------------------------------------------

now for the second one

you have the height of the pyramid, but not the slant-height, which will be the height of the triangular face

thus... check the picture below, using the pythagorean theorem, you can just get the height of the triangular face then

now, the base of the triangles are all 22
and the base of the pyramid is a 22x22 square

---------------------------------------------------------------------------------------

now, for the third one

same as the first one really, you have both, the base and slant-height, which will be the height of the triangular face

so, four triangles with a base of 6 and a height of 9, and a square on the base of 6x6

so... get the area of all four triangles, and the squarish base, sum them up, and that's the surface area of the pyramid

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Complete the table of values
Agata [3.3K]
Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify! 

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, 3^{-2} =
\frac{1}{3^{2} }.
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, (\frac{3}{4}) ^{3}  =  \frac{ 3^{3} }{4^{3} }.
3) The zero exponent rule<span> says that any number raised to zero is 1. For example, 3^{0} = 1.
</span>

Back to the Problem:
Problem 1 
The x-values are in the left column. The title of the right column tells you that the function is y =  4^{-x}. The x-values are:
<span>1) x = 0
</span>Plug this into y = 4^{-x} to find letter a:
y = 4^{-x}\\&#10;y = 4^{-0}\\&#10;y = 4^{0}\\&#10;y = 1
<span>
2) x = 2
</span>Plug this into y = 4^{-x} to find letter b:
y = 4^{-x}\\ &#10;y = 4^{-2}\\ &#10;y =  \frac{1}{4^{2}} \\  &#10;y= \frac{1}{16}
<span>
3) x = 4
</span>Plug this into y = 4^{-x} to find letter c:
y = 4^{-x}\\ &#10;y = 4^{-4}\\ &#10;y =  \frac{1}{4^{4}} \\  &#10;y= \frac{1}{256}
<span>

Problem 2
</span>The x-values are in the left column. The title of the right column tells you that the function is y =  (\frac{2}{3})^x. The x-values are:
<span>1) x = 0
</span>Plug this into y = (\frac{2}{3})^x to find letter d:
y = (\frac{2}{3})^x\\&#10;y = (\frac{2}{3})^0\\&#10;y = 1
<span>
2) x = 2
</span>Plug this into y = (\frac{2}{3})^x to find letter e:
y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\&#10;y =  \frac{4}{9}
<span>
3) x = 4
</span>Plug this into y = (\frac{2}{3})^x to find letter f:
y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}
<span>
-------

Answers: 
a = 1
b = </span>\frac{1}{16}<span>
c = </span>\frac{1}{256}
d = 1
e = \frac{4}{9}
f = \frac{16}{81}
5 0
3 years ago
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