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Olin [163]
4 years ago
15

Which of the following is true about the expression square root of 3*2

Mathematics
1 answer:
baherus [9]4 years ago
7 0

Answer:

B

Step-by-step explanation:

\sqrt{3}*2=3.464101615137755‬ so it is equal to an irrational number.

2 is rational, and \sqrt{3} is irrational, because it is equal to 1.732050807568877‬

Hope this helps you. Please mark brainliest! Have a nice day!

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If tanA = 1/2 then evaluate cos A/sinA +sinA /1+cosA​
Deffense [45]

Answer:

√5.

Step-by-step explanation:

Tan A = 1/2 means that the right triangle containing angle A has legs of length 1 and 2 units. So the hypotenuse  = √(1^2 + 2^2) = √5  (using the Pythagoras theorem). The side opposite to < A = 1 unit and the adjacent side = 2 (as tan = opposite / adjacent).

so cos A =  adjacent / hypotenuse =   2/√5.

and sin A = opposite / hypotenuse = 1 / √5

cos A / sin A = 2/√5 /  1/ √5 =  2.

sin A / (1 + cos A) =  1/√5 (1 + 2/ √5)

=  1 / √5 ( (√5 + 2) /√5)

= 1 / (√5 + 2)

So the answer is:

2  +   1 /(√5 + 2).

We can simplify it further by multiplying top and bottom of the fraction by the complement of √5 + 2 which is √5 - 2.

2 + 1 / (√5 + 2)

= 2(√5 + 2) + 1 / (√5 + 2 )

= { 2(√5 + 2) + 1 } /  (√5 + 2)

Multiplying this by  √5 - 2 / √5 - 2 we get:

(2(5 - 4) + √5 - 2) /  (5 -4)

=  2 + √5 - 2 / 1

= √5.

3 0
3 years ago
What is the sum of (–2.1x + 3.7) and (5 + 4.9x)?
adoni [48]
Combine like-terms
-2.1x and 4.9x = 2.8x
3.7 and 5 = 8.7
2.8x + 8.7 is your final answer.
3 0
4 years ago
Read 2 more answers
Help last question I’ll tell y’all my score!
Rudiy27

Answer:

15

Step-by-step explanation:

graph shows

postcards=15

shirts=10

pencils= 40

we're going to add postcards and shirts THEN comparing it to pencils

postcards and shirts=15+10= 25

pencils= 40

answer=40-25=15

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4 years ago
Which ordered pair is the best estimate for the solution of the system of equations?
gtnhenbr [62]
The ordered pair is 5/2
7 0
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What's the greatest number of pill bugs found
Nimfa-mama [501]
Yea do you have anything that can show for the question then I might help with your questions:)
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