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3 years ago

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How do the number of faces, vertices, and edges of a cube compare to the number of faces, vertices, and edges of a tetrahedron?

A cube has more faces than a tetrahedron. A cube has more vertices than a tetrahedron. A cube has more edges than a tetrahedron.

1 answer:

liq [111]3 years ago

4 0

Answer:

it is 2, 4 and 6

Step-by-step explanation:

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Zolol [24]

Answer:

90

Step-by-step explanation:

In a box plot, the point farthest to the line is the greatest or maximum number of the problem(in this case E). You look for the greatest number out of them all and that would be your answer(90).

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2 years ago

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If you divide the age of the first President Bush when he was inaugurated by 2 and add 14 years, you get the age of President Cl

konstantin123 [22]

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64

Step-by-step explanation:

32+14=46

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3 years ago

If tanA = 3cotA, show that, secA = 2

zysi [14]

Answer:

proved

Step-by-step explanation:

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3 years ago

A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between

sattari [20]

Answer:

a)

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) 0.09

Step-by-step explanation:

We are given the following in the question:

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)$

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) rate of increase after one day.

We put t = 1

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09$

The rate of increase after 1 day is 0.09

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3 years ago

Math Lab Review Work

umka2103 [35]

Answer:

I think it C because I know that Nalia isn't correct so it's not D or B.

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3 years ago