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Vesna [10]
3 years ago
9

Could a polyhedron exist with the given number of faces, vertices, and edges? Drag yes or no to each combination

Mathematics
2 answers:
Bogdan [553]3 years ago
7 0

Answer:

no

NO

NO

Step-by-step explanation:

good luck have a nice day

DanielleElmas [232]3 years ago
5 0

Answer:

(a) No

(b) No

(c) No

Step-by-step explanation:

Given

See attachment

Required

Select Yes or No for each

To do this, we make use of Euler's formula

F + V - E = 2

Where

F \to Faces; V \to Vertices; E \to Edges

(a):\ Faces = 8; Vertices = 12; Edges = 6

Using: F + V - E = 2

8 + 12 - 6 = 2

14 = 2

<em>The above equality is false. Hence, (a) does not exist</em>

(b):\ Faces = 6; Vertices = 6; Edges = 4

Using: F + V - E = 2

6 + 6 - 4 = 2

8 = 2

<em>The above equality is false. Hence, (b) does not exist</em>

<em />

<em />(c):\ Faces = 20; Vertices = 30; Edges = 12<em />

Using: F + V - E = 2

20 + 30 - 12 = 2

38 = 2

<em>The above equality is false. Hence, (c) does not exist</em>

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Vesna [10]

Answer:

C

Step-by-step explanation:

C, because the second line keeps adding 6 to it and, the third line keeps adding 7.

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2 years ago
This Venn diagram has three regions.
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Answer:

B

Step-by-step explanation:

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3 years ago
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3 years ago
A biologist studied how cells grow by adding liquid protein to cell samples. She reported in her study that she divided 3/5 of a
r-ruslan [8.4K]

Answer: \dfrac{1}{5}\text{ of an ounce.}

Step-by-step explanation:

Given : A biologist studied how cells grow by adding liquid protein to cell samples.

She divided \dfrac{3}{5} of an ounce of liquid protein evenly among 3 cell samples.

Here , total quantity of liquid protein = \dfrac{3}{5} of an ounce

Total cell samples = 3

Now , Quantity of liquid protein given by her to each cell = (total quantity of liquid protein ) ÷ (3)

= \dfrac{3}{5}\times\dfrac{1}{3}=\dfrac{1}{5}\text{ of an ounce} .

Hence, Quantity of liquid protein given by her to each cell =  \dfrac{1}{5}\text{ of an ounce.}

4 0
3 years ago
15. Explain why we need to specify 0 &lt; b &lt; 1 and b &gt; 1 as valid values for the base b in the expression logb(xx).
damaskus [11]

Answer:

The base (b) has to be positive and different of 1. The logarithm is the inverse of exponential, so:

logb(a) = x ⇒ a = bˣ

So, for b = 0 ⇒ 0ˣ = a

And there is impossible, "a" only could be 0.

For b = 1 ⇒ 1ˣ = a

And the same thing would happen, the logarithming would be to be 1, and the function will be extremally restricted.

For b<0, then the expression a = bˣ will be also restricted, and will not represent all values of a.

So, 0<b<1 and b >1.

3 0
3 years ago
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