An arithmetic

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

8 0

Answer:

an = 7.5 - 5.5(n - 1).

Step-by-step explanation:

The common difference d = -5.5.

The first term is given as 7.5.

Explicit formula is

an = a1 + (n - 1)d

= 7.5 + (n - 1)-5.5

= 7.5 - 5.5(n - 1)

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All the edges of a regular square pyramid have length 8. find the Volume of the pyramid, Lateral area of the pyramid and the Tot

RUDIKE [14]

Answer:

V = 120.6796

LA = 110.8513

SA = 174.8513

Step-by-step explanation:

Let me know if you need more of an explanation for any of these

First we need the height, one way to find that out is to use the diagonal of the and one of the diagonal edges of the pyramid. Specifically we will need half of the diagonal of the square. We will call the diagonal of the square d.

$8^2+8^2=d^2\\d=\sqrt{64+64} \\d=\sqrt{128}$

Gonna leave it like that for ease of writing. So then half of the diagonal is $\frac{\sqrt{128}}{2}$

The height, h, will be $(\frac{d}{2})^2+h^2 = 8^2$ . Plugging everything in gets us $h=\sqrt{32}$

last, we need to find slant height, which we will call s.

$(\frac{8}{2})^2+h^2 = s^2\\s=\sqrt{48}$

s is also the height of the triangles for purposes of finding area. Now for the volume and areas.

Volume is area of the base times height divided by 3, so $V=8^2\frac{\sqrt{32} }{3} = \frac{64\sqrt{48} }{3} = 120.6796$

Lateral area is the sum of the four areas of the triangles, and each of those are $\frac{8*h}{2}$ so the whole lateral area is 4 times that. So we get $4\frac{8\sqrt{32} }{2} =110.8513$