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Ad libitum [116K]
2 years ago
7

Simplify (x^2)^3*5x/6x^2*15x^3

Mathematics
1 answer:
nasty-shy [4]2 years ago
8 0

When we Simplify [(x^2)^3 × 5x] / [6x^2 × 15x^3], the result obtained is (1/18)x^2

<h3>Data obtained from the question</h3>
  • [(x^2)^3 × 5x] / [6x^2 × 15x^3]
  • Simplification =?

<h3>How to simplify [(x^2)^3 × 5x] / [6x^2 × 15x^3]</h3>

[(x^2)^3 × 5x] / [6x^2 × 15x^3]

Recall

(M^a)^b = M^ab

Thus,

(x^2)^3 = x^6

  • [(x^2)^3 × 5x] / [6x^2 × 15x^3] = [x^6 × 5x] / [6x^2 × 15x^3]

Recall

M^a × M^b = M^(a+b)

Thus,

x^6 × 5x = 5x^(6 + 1) = 5x^7

6x^2 × 15x^3] = (6×15)x^(2 + 3) = 90x^5

  • [x^6 × 5x] / [6x^2 × 15x^3] = 5x^7 / 90x^5

Recall

M^a ÷ M^b = M^(a - b)

Thus,

5x^7 ÷ 90x^5 = (5÷90)x^(7 - 5) = (1/18)x^2

Therefore,

  • [(x^2)^3 × 5x] / [6x^2 × 15x^3] = (1/18)x^2

Learn more about algebra:

brainly.com/question/2768008

#SPJ1

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If (3-4i) (x+yi) = 1 + 0i. The value of x and y is
Black_prince [1.1K]

Step-by-step explanation:

\textsf{\large{\underline{Solution}:}}

<h3><u>Given That:</u></h3>

\rm: \longmapsto (3 - 4i)(x + iy) = 1 + 0i

\rm: \longmapsto 3(x + iy)  - 4i(x + iy)= 1 + 0i

\rm: \longmapsto 3x + 3iy - 4ix +4y= 1 + 0i

<u>On rearranging the terms, we get:</u>

\rm: \longmapsto (3x  + 4y)+ (3y - 4x)i =  1 + 0i

<u>Comparing both sides, we get:</u>

\rm: \longmapsto 3x  + 4y = 1 --- (i)

\rm: \longmapsto 3y - 4x = 0 --- (ii)

<u>Multiplying (i) by 4, we get:</u>

\rm: \longmapsto 12x  + 16y = 4--- (iii)

<u>Multiplying (ii) by 3, we get:</u>

\rm: \longmapsto 9y - 12x = 0 --- (iv)

<u>Adding equations (iii) and (iv), we get:</u>

\rm: \longmapsto25y = 4

\rm: \longmapsto y =\dfrac{4}{25}

<u>From (ii), we get:</u>

\rm: \longmapsto 3y - 4x = 0

\rm: \longmapsto 3y  = 4x

\rm: \longmapsto x =  \dfrac{3}{4} y

\rm: \longmapsto x =  \dfrac{3}{4}  \times  \dfrac{4}{25}

\rm: \longmapsto x =  \dfrac{3}{25}

<u>Therefore:</u>

\rm: \longmapsto (x,y)= \bigg(  \dfrac{3}{25}, \dfrac{4}{25} \bigg)

★ <u>Which is our required answer.</u>

\textsf{\large{\underline{More To Know}:}}

\rm1.\  i^{4n} = 1

\rm2. \ i^{4n+1} = i

\rm3.\  i^{4n+2} = -1

\rm4.\ i^{4n+3} = -i

4 0
3 years ago
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