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Elina [12.6K]
3 years ago
14

Find the value of x:2x+3=9​

Mathematics
2 answers:
larisa [96]3 years ago
7 0

Answer:

x=3

Step-by-step explanation:

2x+3=9 (subtract 3 from both sides)

2x=6 (divide both sides by 2)

x=3

Inga [223]3 years ago
4 0

Answer:

Hello There!!

Step-by-step explanation:

2x+3=9 (-3 from both sides)

-3 -3

2x=6 (divide by 2)

÷2 ÷2

x=3 (6÷2=3)

hope this helps,have a great day!!

~Pinky~

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Step-by-step explanation:

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Step-by-step explanation:

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5 0
2 years ago
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Read 2 more answers
Sin2x-sin2xcos2x=sin4x
yaroslaw [1]

It looks like the given equation is

sin(2x) - sin(2x) cos(2x) = sin(4x)

Recall the double angle identity for sine:

sin(2x) = 2 sin(x) cos(x)

which lets us rewrite the equation as

sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)

Move everything over to one side and factorize:

sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0

sin(2x) - 3 sin(2x) cos(2x) = 0

sin(2x) (1 - 3 cos(2x)) = 0

Then we have two families of solutions,

sin(2x) = 0   or   1 - 3 cos(2x) = 0

sin(2x) = 0   or   cos(2x) = 1/3

[2x = arcsin(0) + 2nπ   or   2x = π - arcsin(0) + 2nπ]

… … …   or   [2x = arccos(1/3) + 2nπ   or   2x = -arccos(1/3) + 2nπ]

(where n is any integer)

[2x = 2nπ   or   2x = π + 2nπ]

… … …   or   [2x = arccos(1/3) + 2nπ   or   2x = -arccos(1/3) + 2nπ]

[x = nπ   or   x = π/2 + nπ]

… … …   or   [x = 1/2 arccos(1/3) + nπ   or   x = -1/2 arccos(1/3) + nπ]

7 0
2 years ago
1. Write a paragraph proof for the following conjecture.
Lesechka [4]

QS bisects the ^PQR^ :

So -----¢ PQS = SQR = 45°

_________________________________

PQR = PQS + SQR = 45°+ 45° = 90 °

Q^ = 90° ----¢ So the PQR triangle is a

right triangle.

4 0
3 years ago
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