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Alenkinab [10]
3 years ago
12

The base of a ladder is placed 6 ft from the bottom of the house. The ladder touches the

Mathematics
2 answers:
sammy [17]3 years ago
4 0

Answer:

10 ft

Step-by-step explanation:

Inessa [10]3 years ago
4 0

Answer:

Step-by-step explanation:

Draw a picture. The ladder is the hypotenuse of a right triangle with legs of length 6 ft and 8 ft.

length of ladder = √(6²+8²) = 10 ft

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agasfer [191]
<span><span>n<span>x4/</span></span>5</span>=<span>3/<span>4

</span></span><span><span><span><span>1/5</span><span>n<span>x^4</span></span></span><span><span>x^4/</span>5</span></span>=<span><span>3/4</span><span><span>x^4/</span>5</span></span></span><span>
Answer is n=
<span>15/<span>4<span>x<span>4</span></span></span></span></span>
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3 years ago
Triangle has sides that measure 33 cm, 65 cm, and 56 cm. Is it a right triangle? Explain
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Yes, by Pythagoras: Larguest side² = other side² + other side² (65²=33²+56²).






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3 years ago
Easy work! Please click on the photo and do 5. 10. 14. And 16.Also it’s is easy but only for other people not me just for people
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branliyist please

Step-by-step explanation:

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)

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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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Firdavs [7]

Answer:

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Step-by-step explanation:

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