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katovenus [111]
3 years ago
12

Could I get some help in here​

Mathematics
1 answer:
IgorC [24]3 years ago
3 0

Answer:

  98 ft²

Step-by-step explanation:

There are a couple of ways you can think about this one. Perhaps easiest is to treat it as a square with a triangle cut out of it. The cutout triangle has a base (across the top) of 14 ft and a height of 14 ft, so its area is ...

  A = (1/2)(14 ft)(14 ft) = 98 ft²

Of course the area of the square from which it is cut is ...

  A = (14 ft)² = 196 ft²

So, the net area of the two triangles shown is ...

  A = (196 ft²) - (98 ft²) = 98 ft²

_____

Another way to work this problem is to attack it directly. Let the base of the left triangle be x. Then the base of the right triangle is 14-x, and their total area is ...

  A = A1 + A2 = (1/2)(x ft)(14 ft) + (1/2)((14-x) ft)(14 ft)

We can factor out 7 ft to get ...

  A = (7 ft)(x ft + (14 -x) ft)

  A = (7 ft)(14 ft) = 98 ft²

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77.  \cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} xProved

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Step-by-step explanation:

77. Left hand side

= \cot^{6} x

= \cot^{4} x \times \cot^{2} x

= \cot^{4}x [\csc^{2}x - 1]  

{Since we know, \csc^{2} x - \cot^{2}x = 1}

= \cot^{4} x \csc^{2}x - \cot^{4} x  

= Right hand side (Proved)

78. Left hand side

= \sec^{4}x \tan^{2} x

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{Since \sec^{2}x - \tan^{2}x = 1}

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= \cos^{3} x\sin^{2} x

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{Since \sin^{2}x + \cos^{2} x = 1}

= [\sin^{2}x - \sin^{4}x] \cos x

= Right hand side

80. Left hand side  

= \sin^{4}x - \cos^{4}x

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{Since \sin^{2}x + \cos^{2} x = 1}

= 1 - 2\cos^{2} x[1 - \cos^{2}x ]

= 1 - 2\cos^{2}x + 2 \cos^{4} x

= Right hand side. (Proved)

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