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2 years ago

7

What is the answer for K ?

2 answers:

Norma-Jean [14]2 years ago

7 0

10/3 = k/9

10 x 9 = 3 x k

90 = 3 k

k = 90/3

k = 30

Luda [366]2 years ago

5 0

You need to do a cross multiple first and then we’ll get 3K=90 so they just want K so 90/3 and the answer is 30

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3 is subtracted from a number.<br> What is the algebraic expression?

Stels [109]

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Step-by-step explanation:

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3 years ago

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Lelechka [254]

Answer:

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

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

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

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79. Left hand side

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

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

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

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

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80. Left hand side

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

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

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3 years ago

Write the expression in repeated multiplication form. Then write the expression as a power. (-8)^3*(-8)^4 The expression in repe

Makovka662 [10]

Given:

The expression is

$(-8)^3\cdot (-8)^4$

To find:

The expression in repeated multiplication form and then write the expression as a power.

Solution:

We have,

$(-8)^3\cdot (-8)^4$

The repeated multiplication form of this expression is

$=[(-8)\cdot (-8)\cdot (-8)]\cdot [(-8)\cdot (-8)\cdot (-8)\cdot (-8)]$

$=(-8)\cdot (-8)\cdot (-8)\cdot (-8)\cdot (-8)\cdot (-8)\cdot (-8)$

Clearly, (-8) is multiplied seven times by itself. So,

$=(-8)^7$

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2 years ago

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Answer:

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2 years ago

suter [353]

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y=5x-2

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3 years ago