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3 years ago

Rewrite x5/6 divided by x1/6 in simplest radical form

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

5 0

5/6 and 1/6 are exponents to X. The X’s are divided by each other. does that make sense?

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Which conclusion of the following statement must always be true?

ad-work [718]

Answer:

option B. they are complementary

Step-by-step explanation:

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3 years ago

Ja'kniya and Alayjah are shopping at the Apple Store. Ja'kniya bought 4 screensavers and 3 Otterboxes for $29. Alayjah bought 6

Katarina [22]

Answer: Each screensaver costs $2 and each Otterbox costs $7.

Step-by-step explanation:

Let x = Cost of each screensaver and y = cost of each Otterbox.

As per given, 4x+3y=29 (i)

6x +2y = 26 (ii)

Divide both sides of (i) by 2 and the multiply it by 3 on sides , we get

9x+3y=39 (iii)

Eliminate (i) from (iii) , we get

5x = 10

⇒ x = 2 [Divide both sides by 5]

Put value of x in (i), we get

4(2)+3y=29

⇒ 8+3y= 29

⇒ 3y= 21

⇒ y =7 [Divide both sides by 7]

Hence, Each screensaver costs $2 and each Otterbox costs $7.

7 0

3 years ago

Help me please!!!!!!!

yawa3891 [41]

The answer would be B

7 0

3 years ago

The sides of a triangle are x, x+7, and 2x-8. If the perimeter of the triangle is 47, what is x

kirza4 [7]

just like the previous one, we sum the sides to get the perimeter.

$\bf (x)+(x+7)+(2x-8)=47\implies 4x-1=47\implies 4x=48\\\\\\x=\cfrac{48}{4}\implies x=12$

8 0

3 years ago

Please help me with these, oh sweet jesus

Lelechka [254]

Answer:

77. $\cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} x$ 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 $\sec^{2}x - \tan^{2}x = 1$ }

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

= 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