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yaroslaw [1]
3 years ago
8

How do you evaluate a logarithm

Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

If you want to solve a logarithm, you can rewrite it in exponential form and solve it that way

Step-by-step explanation:

i don't know if this will help but i tried

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What is the volume, in cubic in, of a cylinder with a height of 8 in and a base radius of
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Answer:

V=100.5

Step-by-step explanation:

hopes this helps

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2 years ago
15.7/6.28 what is the answer how do you do it
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Simply divide 15.7 by 6.28 to get 2.5 :) easy peasy pumpkin pie
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3 years ago
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drek231 [11]

Answer:                                                                                

When a 2-D circle is rotated around a line that it does not touch. What is the name of the resulting figure?  

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8 0
4 years ago
Suppose you have 18 objects (10 of type A, 5 of type B, and 3 of type C). Objects of type A are indistinguishable from each othe
Paladinen [302]

Answer:

\binom{18}{5}= 8568

Step-by-step explanation:

Note that we have in total 18 items. Even though we are given information regarding the amounts of items per type, the general question asks the total number of ways in which you can pick 5 out of the 18 objects, without any restriction on the type of chosen items. Therefore, the information regarding the type is unnecessary to solve the problem.

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5 0
3 years ago
Read 2 more answers
Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select
Blizzard [7]

Answer:

Probability is:   $ \frac{\textbf{13}}{\textbf{51}} $

Step-by-step explanation:

From a deck of 52 cards there are 26 black cards. (Spades and Clubs).

Also, there are 26 red cards. (Hearts and Diamonds).

First, we determine the probability of drawing a black card.

P(drawing a black card) = $ \frac{number \hspace{1mm} of  \hspace{1mm} black  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm} of  \hspace{1mm} cards} $  $ = \frac{26}{52} = \frac{\textbf{1}}{\textbf{2}} $

Now, since we don't replace the drawn card, there are only 51 cards.

But the number of red cards is still 26,

∴ P(drawing a red card) = $ \frac{number  \hspace{1mm} of  \hspace{1mm} red  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm}of  \hspace{1mm} cards} $  $ = \frac{26}{51}  $

Now, the probability of both black and red card = $ \frac{1}{2} \times \frac{26}{51} $

$ = \frac{\textbf{13}}{\textbf{51}} $

Hence, the answer.

5 0
3 years ago
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