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Lelu [443]
2 years ago
10

Can someone please help me with these 7 questions please?

Mathematics
1 answer:
yarga [219]2 years ago
3 0

(1)\ (-xy)^3(xz)

Expand

(-xy)^3(xz) = (-x)^3* y^3*(xz)

(-xy)^3(xz) = -x^3* y^3*xz

Rewrite as:

(-xy)^3(xz) = -x^3*x* y^3*z

Apply law of indices

(-xy)^3(xz) = -x^4y^3z

(2)\ (\frac{1}{3}mn^{-4})^2

Expand

(\frac{1}{3}mn^{-4})^2 =(\frac{1}{3})^2m^2n^{-4*2}

(\frac{1}{3}mn^{-4})^2 =\frac{1}{9}m^2n^{-8

(3)\ (\frac{1}{5x^4})^{-2}

Apply negative power rule of indices

(\frac{1}{5x^4})^{-2}= (5x^4)^2

Expand

(\frac{1}{5x^4})^{-2}= 5^2x^{4*2}

(\frac{1}{5x^4})^{-2}= 25x^{8

(4)\ -x(2x^2 - 4x) - 6x^2

Expand

-x(2x^2 - 4x) - 6x^2 = -2x^3 + 4x^2 - 6x^2

Evaluate like terms

-x(2x^2 - 4x) - 6x^2 = -2x^3 -2x^2

Factor out x^2

-x(2x^2 - 4x) - 6x^2 = (-2x-2)x^2

Factor out -2

-x(2x^2 - 4x) - 6x^2 = -2(x+1)x^2

(5)\ \sqrt{\frac{4y}{3y^2}}

Divide by y

\sqrt{\frac{4y}{3y^2}} = \sqrt{\frac{4}{3y}}

Split

\sqrt{\frac{4y}{3y^2}} = \frac{\sqrt{4}}{\sqrt{3y}}

\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}}

Rationalize

\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}} * \frac{\sqrt{3y}}{\sqrt{3y}}

\sqrt{\frac{4y}{3y^2}} = \frac{2\sqrt{3y}}{3y}

(6)\ \frac{8}{3 + \sqrt 3}

Rationalize

\frac{8}{3 + \sqrt 3} = \frac{3 - \sqrt 3}{3 - \sqrt 3}

\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{(3 + \sqrt 3)(3 - \sqrt 3)}

Apply different of two squares to the denominator

\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{3^2 - (\sqrt 3)^2}

\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{9 - 3}

\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{6}

Simplify

\frac{8}{3 + \sqrt 3} = \frac{4(3 - \sqrt 3)}{3}

(7)\ \sqrt{40} - \sqrt{10} + \sqrt{90}

Expand

\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4*10} - \sqrt{10} + \sqrt{9*10}

Split

\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4}*\sqrt{10} - \sqrt{10} + \sqrt{9}*\sqrt{10}

Evaluate all roots

\sqrt{40} - \sqrt{10} + \sqrt{90} =2*\sqrt{10} - \sqrt{10} + 3*\sqrt{10}

\sqrt{40} - \sqrt{10} + \sqrt{90} =2\sqrt{10} - \sqrt{10} + 3\sqrt{10}

\sqrt{40} - \sqrt{10} + \sqrt{90} =4\sqrt{10}

(8)\ \frac{r^2 + r - 6}{r^2 + 4r -12}

Expand

\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r^2 + 3r-2r - 6}{r^2 + 6r-2r -12}

Factorize each

\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r(r + 3)-2(r + 3)}{r(r + 6)-2(r +6)}

Factor out (r+3) in the numerator and (r + 6) in the denominator

\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{(r -2)(r + 3)}{(r - 2)(r +6)}

Cancel out r - 2

\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r + 3}{r +6}

(9)\ \frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14}

Cancel out x

\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 5x - 14}

Expand the numerator of the 2nd fraction

\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 7x+2x - 14}

Factorize

\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x(x - 7)+2(x - 7)}

Factor out x - 7

\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{(x + 2)(x - 7)}

Factor out 4 from 4x + 8

\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4(x + 2)}{x} \cdot \frac{1}{(x + 2)(x - 7)}

Cancel out x + 2

\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x} \cdot \frac{1}{(x - 7)}

\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x(x - 7)}

(10)\ (3x^3 + 15x^2 -21x) \div 3x

Factorize

(3x^3 + 15x^2 -21x) \div 3x = 3x(x^2 + 5x -7) \div 3x

Cancel out 3x

(3x^3 + 15x^2 -21x) \div 3x = x^2 + 5x -7

(11)\ \frac{m}{6m + 6} - \frac{1}{m+1}

Take LCM

\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m(m + 1) - 1(6m + 6)}{(6m + 6)(m + 1)}

Expand

\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 + m- 6m - 6}{(6m + 6)(m + 1)}

\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 - 5m - 6}{(6m + 6)(m + 1)}

(12)\ \frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}}

Rewrite as:

\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} \div \frac{2}{y^2 - 9}

Express as multiplication

\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 9}{2}

Express y^2 - 9 as y^2 - 3^2

\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 3^2}{2}

Express as difference of two squares

\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{(y - 3)(y+3)}{2}

\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{1} * \frac{(y+3)}{2}

\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{y+3}{2}

Read more at:

brainly.com/question/4372544

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