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mixer [17]
3 years ago
11

Work out 5^-2 x 3 square root 8

Mathematics
1 answer:
Serjik [45]3 years ago
5 0

Answer:

6root(2)/25

Step-by-step explanation:

5^-2 = 1/25

3(root(8)) = 6(root(2))

so you get 6(root(2))(1/25)

= 6(root(2))/25

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jek_recluse [69]

Answer:

SSS and SAS

Step-by-step explanation:

Law of cosines: The law of cosines is used for calculating one side of a triangle when the angle opposite and the other two sides are known.

If the information about a triangle according to SSS, SAS and ASA is given, then we will immediately use SSS and SAS  by using the law of cosines to find one of the remaining measures because Law of cosines is applied if we know one angle opposite and the other two sides.

Also, in ASA, we know two angles and one side, thus we cannot use Law of cosines in this, rather Law of sines can be accurately use in this case.

if you have two sides, encroaching an angle, then the Law of Cosines can be used to find a missing side, thus SAS will work fine for that, now, if you have no angles given, but you know all sides, you can use the Law of Cosines as well, by solving it for the angle, and get the angles, on which case, SSS will do... .now as far as ASA, if you have two angles and one side known, then... that's not very workable for the law of cosines

please mark me brainliest :)

3 0
3 years ago
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Rationalize the denominator of square root of negative 9 over open parentheses 4 minus 7 i close parentheses minus open parenthe
svetlana [45]
First we combine the parenthesis to get 9/(-2-i)  then we multiply both the numerator and the denomenator by the conjugate of -2-i (whic is -2+i) to get (-18+9i)/5
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3 years ago
How do you write 5,621 divided by 23 word problem
Alona [7]
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2 years ago
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Help me with 26 ASAP <br> Please show work<br> Random Answers will be moderated! Thanks :)
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4 0
3 years ago
What the questions are asking help ASAP please
Rudiy27

Answer:

Step-by-step explanation:

A scalar is a constant value that is multiplied throughout a matrix.

e.g.

In number 1 the set up would look like this

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The solution to #1 would be

M = \left[\begin{array}{ccc}9&-3&15\\6&3&-12\\-18&9&6\end{array}\right]

3 0
3 years ago
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