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Westkost [7]
2 years ago
8

PPLLLLLLLLLLLLLLEEEEEEEEEEEAAAAAAAAAASSSSSSSSSSSSSEEEEEEEEEE Help

Mathematics
1 answer:
STALIN [3.7K]2 years ago
3 0

Answer:

its 35.

Step-by-step explanation:

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Please answer quick!
marysya [2.9K]

Answer:

The correct option is B

(-1)(1/2)(-1)(1)

Step-by-step explanation:

First thing to notice is that there is are two brackets with negative values, which means that the result must also be positive (or have two brackets with negative values).

Looking at the options, we can screen out options A and D, they have three brackets with negative values, and can't be chosen.

It is now between options B and C.

Option B, by inspection is simply 1/2

Option C is 12/4 = 3

The problem itself is 12/35

12/24 = 1/2

12/36 = 1/3

Since 12/35 is about 1/3, it is closer to 1/2 than 3, so 1/2 is the best option.

6 0
2 years ago
Please Help ! Dilations
Darina [25.2K]

Answer:

Answer is C.

Step-by-step explanation:


4 0
3 years ago
What is the approximate volume of the cone?
Vlad [161]
Volume=1/3π<span>×</span>r²×h^(Applies for any "right cone")^
Therefore your answer would be 
A.1206 cm³

8 0
2 years ago
Read 2 more answers
George says that 4+ a +a+a+b+b=24ab
antoniya [11.8K]

Answer:

3a+2b+4

Step-by-step explanation:

24ab indicates that there is 24 of unit ab which there is not. there is 4+ 3a+2b

4 0
2 years ago
Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u
Tamiku [17]

Answer:

Step-by-step explanation:

15. \frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx}   \\

16.  \frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }

=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2}  \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\

=> The clause is correct

17. Do the same (16)

18. \frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx

19.

\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\

=> The clause is correct

20. Do the same (18)

21. Left = \frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA}  } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}

Right = cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA}   \\

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

5 0
3 years ago
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