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prisoha [69]
3 years ago
7

PLEASE HELP EMERGENCY !! PLS PLS

Mathematics
2 answers:
Elena-2011 [213]3 years ago
6 0

Answer:

4

Step-by-step explanation:

tatyana61 [14]3 years ago
4 0
Call 911 then but the answer is $4 because 12/3 is 4
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Verify identity: <br><br> (sec(x)-csc(x))/(sec(x)+csc(x))=(tan(x)-1)/(tan(x)+1)
Nikitich [7]
So hmmm let's do the left-hand-side first

\bf \cfrac{sec(x)-csc(x)}{sec(x)+csc(x)}\implies \cfrac{\frac{1}{cos(x)}-\frac{1}{sin(x)}}{\frac{1}{cos(x)}+\frac{1}{sin(x)}}\implies &#10;\cfrac{\frac{sin(x)-cos(x)}{cos(x)sin(x)}}{\frac{sin(x)+cos(x)}{cos(x)sin(x)}}&#10;\\\\\\&#10;\cfrac{sin(x)-cos(x)}{cos(x)sin(x)}\cdot \cfrac{cos(x)sin(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}

now, let's do the right-hand-side then  

\bf \cfrac{tan(x)-1}{tan(x)+1}\implies \cfrac{\frac{sin(x)}{cos(x)}-1}{\frac{sin(x)}{cos(x)}+1}\implies \cfrac{\frac{sin(x)-cos(x)}{cos(x)}}{\frac{sin(x)+cos(x)}{cos(x)}}&#10;\\\\\\&#10;\cfrac{sin(x)-cos(x)}{cos(x)}\cdot \cfrac{cos(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}

7 0
2 years ago
11.45 x 3.7 = ? <br><br>(pa help) ​
Stels [109]

Answer:

42.365

i used a calculator

7 0
2 years ago
The mean price for used cars is $10,550. A manager of a Kansas City used car dealership reviewed a sample of 50 recent used car
Evgen [1.6K]

Answer:

go to algebra answers.com

Step-by-step explanation:

8 0
2 years ago
Is 0.71428571428 a rational number
Alex787 [66]
No, since it a repeating(terminating) decimal, so it is not a rational number.

5 0
3 years ago
Read 2 more answers
Which expressions are equivalent to this expression?
Lyrx [107]

Answer:

Option B

Option C

Option D

All of these are correct

Step-by-step explanation:

Option B is correct because:

2x + 1 + 2x + 3 = 4x + 4

=> 4x + 4 = 4x + 4

Option C is correct because:

x + 3x + 4 = 4x + 4

=> 4x + 4  = 4x + 4

Option D is correct because:

2(2x + 2) = 4x + 4

=> 4x + 4 = 4x + 4

Option A is incorrect because:

4(x + 4) = 4x + 4

=> 4x + 16 ≠ 4x + 4

7 0
3 years ago
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