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Jobisdone [24]
3 years ago
12

Solve the inequality 2x - 3 Need help.... plz

Mathematics
2 answers:
timama [110]3 years ago
6 0
2x - 3 = 0
2x = 0 + 3
2x = 3
X = 3/2 , 1.5
-BARSIC- [3]3 years ago
5 0

Answer:

x=3/2

Step-by-step explanation:

2x-3=0

2x=0+3

2x=3

x=3/2

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What part are you lost on, do you want me to teach you how to divide them?

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Two expressions are shown below:
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Answer:

They are not equal

Step-by-step explanation:

So, lets start by simplifying the expressions

Expression 1:

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Expression 2:

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We have two different expressions now, they don't seem to be equivalent, since in the first one, we have 4 x s + 2, but in the second one, it is s x s 2 times added with h x h.

If you have any questions, let me know in the comments. If you could mark this answer as the brainliest, I would very much appreciate it! :D

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2 years ago
(cosxtanx-tanx+2cosx-2)/tanx+2=cosx-1<br> can someone show me how to prove this?
uysha [10]

Answer:

See detail below.

Step-by-step explanation:

A word of caution before getting to the actual problem: I believe there is an important set of brackets missing in the original post. The expression on the left hand side should be:

(cosxtanx-tanx+2cosx-2)/(tanx+2)

Without the brackets, it is left unclear whether the denominator is just tanx or tanx+2. I recommend to use brackets wherever any doubt could arise.

Now to the actual problem: \we can make the following transformations on the left hand side:

\frac{\cos x \tan x + 2\cos x -2}{\tan x +2}=\frac{\cos x \frac{\sin x}{\cos x}  + 2\cos x -2}{\frac{\sin x +2\sin x}{\cos x} }=\\=\frac{\cos x \sin x - \sin x + 2\cos^2 x - 2\cos x}{\sin x + 2\cos x}=\\=\frac{\cos x \sin x +2\cos^2 x }{\sin x + 2\cos x}-1=\\=\cos x -1

which is shown to be the same as the right hand side, which was to be shown.

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