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lana [24]
2 years ago
11

Please can someone help!!!

Mathematics
2 answers:
Scorpion4ik [409]2 years ago
6 0

Answer:

choice #3 (y=4x+2)

(y=mx+b)

m= slope= 4

b= y-intercept= 2

liraira [26]2 years ago
5 0

Answer:

Step-by-step explanation:

y= 4x+2

trust me!!:)

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Please answer this now in two minutes
postnew [5]

Answer:

8

Step-by-step explanation:

1 2 3 4 5 6 7 8 quick math

6 0
3 years ago
Solve for X: <br> 5(2x-12)=140
Mrac [35]

Answer:

x=20

Step-by-step explanation:

10x-60=140

10x=140+60

10x=200

x=20

6 0
3 years ago
Read 2 more answers
The expression 9/5C + 32, where C stands for the temperature in degrees Celsius, is used to convert Celsius to Fahrenheit. If th
Ira Lisetskai [31]
<span>9/5C + 32 =

= 9/5 * 20 + 32

= 36 + 32

= 68

20 C = 68 F
</span>
3 0
3 years ago
What's the value of x?
marshall27 [118]
The value of x is 2.5
6 0
3 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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