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Daniel [21]
3 years ago
14

Complete the point-slope equation of the line through (-8,-1)(−8,−1)left parenthesis, minus, 8, comma, minus, 1, right parenthes

is and (-6,5)(−6,5)left parenthesis, minus, 6, comma, 5, right parenthesis.
Use exact numbers.
y-5=y−5=y, minus, 5, equals
Mathematics
2 answers:
ratelena [41]3 years ago
4 1

Answer:

- y=(−112)x−52. Explanation: (−10,3) and (−8,−8) The slope of the line between A(x1,y1) and B(x2,y2) is: m=y2−y1x2−x1.

Step-by-step explanation:

gbjkngfjfg2 years ago
0 0

DOGSHIT ^ DAT KIDDO IS TRIPPPPPPPY.

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Least to greatest -5, -5.2, 5.5, -5 1/2, -5/2
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Answer: -5 1/2, -5.2, -5, -5/2, 5.5

We can visualize a number line to see where these numbers would rank.

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Find the compound Interest for Rs.18,800 ,calculated for 2 years at 13% rate of Interest compounded annually
Vinvika [58]

For compound interest, the formula is given below:

Amount = P(1+\frac{r}{100} )^{n}

Here, P = 18,800

n = 2

r = 13/100

So, Amount = 18,800(1+\frac{13}{100} )^{2}

18,800(1.13)^{2}

= 18,800 × 1.2769

= 24005.72

Compound Interest = Amount - Principal

Compound Interest = 24005.72 - 18800

= 5205.72

Hence, the compound interest for Rs.18,800, calculated for 2 years at 13% rate of interest compounded annually is Rs.5205.72.


4 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 :))))

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Solve the following compound inequality.
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