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EastWind [94]
3 years ago
9

Simplify: 5 - i8 + 2i32

Mathematics
1 answer:
cricket20 [7]3 years ago
5 0
So when writing non-real numbers you must put it into a+bi format where a is the real part and bi is the non-real part so
if we assume that 2i32=2*32i=64i
then the closed phrase is
5-8i+64i
or
5+56i
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2 years ago
Please help me with these, oh sweet jesus
Lelechka [254]

Answer:

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

78.  \sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ] Proved

79. \cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x Proved.

80. \sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x Proved.

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

= \sec^{2} x [1 + \tan^{2}x] \tan^{2} x  

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

= \sec^{2}x [\tan^{2}x + \tan^{4}x ]

= Right hand side (Proved)

79. Left hand side  

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

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

{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

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

{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)

7 0
3 years ago
Use the diagram to answer the question.
Firlakuza [10]
D.Q (3,0)
I think
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8 0
3 years ago
Brainliest answer
yan [13]

Answer:

1

Step-by-step explanation:

3 0
3 years ago
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What is 8.786 rounded to nearest tenth
pentagon [3]
8.8 because .78 rounded up is .8
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3 years ago
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