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erastova [34]
3 years ago
5

Ebgjwbjhkjfhdsn wmiuehdbnsmkji

Mathematics
2 answers:
posledela3 years ago
5 0

DYK?

— Pretty face is everywhere, but a woman who cheer you up at your weakest and stay after knowing about your flaws and bad sides are rare, so you better keep her dude!.

#moodchallenge

natali 33 [55]3 years ago
4 0

Answer:

im srry what

Step-by-step explanation:

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5. Class 7887 has 2:7 math classes daily. How many math classes do they have
ANTONII [103]

Answer:

56.7

Step-by-step explanation:

2.7 x 21 = 56.7

5 0
3 years ago
I need help this is due tomorrow and I have to write an essay so please help!!!
HACTEHA [7]
2.) 6k - 9
*can't see #3*
4.) -9p + 17
5.) -15b^2 - 1b + 6c
6.) -4j
4 0
3 years ago
I'm in algebra 1 please help i'm lost
german
The answer is #3... 2 and -6! Hope this helped!
5 0
4 years ago
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
3 years ago
Please help me :(<br>NO LINKS PLEASE<br>​
liubo4ka [24]

Answer:

Option A

Step-by-step explanation:

<u>Given equation is:</u>

x^2+1=2x-3

Setting it up as a quadratic equation.

x^2+1-2x+3=0\\\\x^2-2x+4=0

Comparing it with quadratic equation ax^2+bx+c=0, we get:

a = 1 , b = -2 , c = 4

<u>Putting the values in quadratic equation:</u>

\displaystyle x = \frac{-b +-\sqrt{b^2-4ac} }{2a} \\\\x = \frac{-(-2)+-\sqrt{(-2)^2-4(1)(4)}  }{2(1)}

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
6 0
3 years ago
Read 2 more answers
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