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(75c^2d^4f^8)/(15cd^3f^4)

75/15 x c^2d^4f^8 over cd^3f^4

5 x c^2d^4f^8 over cd^3f^4

5c^2-1d^4-3f^8-4

5c^1d^4-3f^8-4

5c^1d^1f^8-4

5c^1d^1f^4

5cd^1f^4

5cdf^4 or option A.
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5 0

3 years ago

Maria is buying new carpet for her bedroom .Her bedroom is in the shape of a square and the length of each side is 12 feet write

Musya8 [376]

Answer:

well just do area, and since it's the same in each side 12×4= 144

3 0

3 years ago

Sec^6x(secxtanx)-sec^4(secxtanx)=sec^5xtan^3x Verify the trigonometric identity

mario62 [17]

I guess you mean

$\sec^6x(\sec x\tan x)-\sec^4x(\sec x\tan x)=\sec^5x\tan^3x$

On the left side, we have a common factor of $\sec^4x(\sec x\tan x)=\sec^5x\tan x$ , so that

$\sec^6x(\sec x\tan x)-\sec^4x(\sec x\tan x)=\sec^5x\tan x(\sec^2x-1)$

Recall that

$\sec^2x=1+\tan^2x$

from which it follows that

$\sec^5x\tan x(\sec^2x-1)=\sec^5x\tan x\tan^2x=\sec^5x\tan^3x$

7 0

3 years ago