Mathematics: Simplify the following expression (4x^-2)^0

Simplify the following expression (4x^-2)^0, №9920247, 25.02.2022 18:26

Mathematics

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is an expression for the length of the pool? a. 2x^2+8x+3 b. 2x^2-8x-3 c. 2x^2-8x+3 d. 2x^2+8-3 2. simplify x/6x-x^2 a. 1/6-x; where x = 0,6 b. 1/6-x; where x=6 c. 1/6; where x=0 d. 1/6 3. simplify -12x4/x4+8x^5 a. -12/1+8x; where x= -1/8 b. -12/1+8x; where x= -1/8,0 4. simplify x+5/ x^2+6x+5 a. 1/x+1; where x= -1 b. 1/x+1; where x=-1, -5 5. simplify x^2-3x-18/x+3 a. x-3 b. x-6; where x= -3 c. x-6; where x= 6 d. 1/x+3; where x= -3 6. simplify 2/3a .

Step-by-step answer

P Answered by PhD

20

Question 1

To find the width of the rectangle, we divide the area by the length
$2x^{3}-29x+12$ ÷ x+4

We use the method of long division to get the answer. The method is shown in the first diagram below

$2x^{2}-8x+3$

Question 2:
$\frac{x}{6x-x^{2} } = \frac{x}{x(6-x)} = \frac{1}{6-x}$

Question 3:
$\frac{-12 x^{4} }{x^{4}+8 x^{5} }= \frac{-12 x^{4} }{ x^{4}(1+8x)}= \frac{-12}{1+8x}$

Question 4:
$\frac{x+5}{x^{2}+6x+5}= \frac{x+5}{(x+1)(x+5)}= \frac{1}{x+1}$

Question 5:
$\frac{x^{2}-3x-18} {x+3}= \frac{(x-6)(x+3)}{x+3}= \frac{x-6}{1}=x-6$

Question 6:
$\frac{2}{3a}$ × $\frac{2}{a^{2}}$ = $\frac{4}{3a^{3} }$ where $a \neq 0$

Question 7: (Question is not written well)
$\frac{x-5}{4x+8}$ × $(12x^{2}+32x+8)$
$\frac{12 x^{3}-28 x^{2} -152x-40 }{4x+8}$
By performing long division we get an answer $3 x^{2} -x-36$ with remainder of 248

Question 8:
$( \frac{x^{2}-16} {x-1})$ ÷ (x+4)

$( \frac{ x^{2}-16 }{x-1})$ × $\frac{1}{x+4}$
$\frac{(x+4)(x-1)}{x-1}$ × $\frac{1}{x+4}$
Cancelling out x+4

we obtain $\frac{x+1}{x-1}$

Question 9:
$\frac{x^{2}+2x+1} {x-2}$ ÷ $\frac{x^{2-1} }{x^{2}-4 }$
$\frac{ x^{2}+2x+1 }{x-2}$ × $\frac{x^{2}-4 }{x^{2}-1}$
Factorise all the quadratic expression gives
$\frac{(x+1)(x+1)}{x-2}$ × $\frac{(x-2)(x+2)}{(x+1)(x-1)}$
Cancelling out (x+1)

and

gives a simplest form
$\frac{(x+1)(x+2)}{x-1}$

Question 10:

$\frac{24 w^{10}+8w^{12} }{4 x^{4} }= \frac{24w^{10} }{4 x^{4} } + \frac{8 w^{12} }{4 x^{4} }$
Cancelling out the constants of each fraction
$\frac{6w^{10} }{x^{4} }+ \frac{2w^{12} }{x^{4}}= \frac{6w^{10}+2w^{12} }{ x^{4}}$

Question 11:

$\frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } = \frac{-2m^{6}(3m^{3}-3m^{2}-8)}{2m^{3} }$
Cancelling $2m^{3}$ gives us the simplified form
$-m^{3}(3m^{3}-3m^{2}-8) = -3m^{6}+3m^{5}+8m^{3}$

Question 12:

$\frac{-4x}{x+7} - \frac{8}{x-7} = \frac{-4x(x-7)-8(x+7)}{(x+7)(x-7)}$
$\frac{-4 x^{2} +28x-8x-56}{(x+7)(X-7)}= \frac{-4 x^{2} +20x-56}{(x+7)(x-7)}$
Factorising the numerator expression
$\frac{(-4x+28)(x-2)}{(x+7)(x-7)} = \frac{-4(x-7)(x-2)}{(x+7)(x-7)}$
Cancelling out x-7

gives the simplified form
$\frac{-4x+8}{x-7}$

Question 13:

$\frac{3}{x-3} - \frac{5}{x-2}= \frac{x3(x-2)-5(x-2)}{y(x-3)(x-2)}$
$\frac{3x-6-5x+15}{(x-3)(x-2)}= \frac{-2x+9}{(x-3)(x-2)}$

Question 14:

$\frac{9}{x-1}- \frac{5}{x+4}= \frac{9(x+4)-5(x-1)}{(x-1)(x+4)}$ $\frac{9x+36-5x+5}{(x-1)(x+4)}= \frac{4x+41}{(x-1)(x+4)}$

Question 15:

$\frac{-3}{x+2}- \frac{(-5)}{x+3}= \frac{-3(x+3)-(-5)(x+2)}{(x+2)(x+3)}$
$\frac{-3x-9+5x+10}{(x+2)(x+3)}= \frac{2x+1}{(x+2)(x+3)}$

Question 16:

$\frac{4}{x}+ \frac{5}{x}=-3$
$\frac{9}{x}=-3$
x=-3

Question 17:

$\frac{1}{3x-6}- \frac{5}{x-2}=12$
$\frac{(x-2)-5(3x-6)}{(3x-6)(x-2)} = \frac{x-2-15x+30}{(3x-6)(x-2)}= \frac{-14x+28}{(3x-6)(x-2)}$

Question 18

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is

Mathematics

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is an expression for the length of the pool? a. 2x^2+8x+3 b. 2x^2-8x-3 c. 2x^2-8x+3 d. 2x^2+8-3 2. simplify x/6x-x^2 a. 1/6-x; where x = 0,6 b. 1/6-x; where x=6 c. 1/6; where x=0 d. 1/6 3. simplify -12x4/x4+8x^5 a. -12/1+8x; where x= -1/8 b. -12/1+8x; where x= -1/8,0 4. simplify x+5/ x^2+6x+5 a. 1/x+1; where x= -1 b. 1/x+1; where x=-1, -5 5. simplify x^2-3x-18/x+3 a. x-3 b. x-6; where x= -3 c. x-6; where x= 6 d. 1/x+3; where x= -3 6. simplify 2/3a .

Step-by-step answer

P Answered by PhD

20

Question 1

To find the width of the rectangle, we divide the area by the length
$2x^{3}-29x+12$ ÷ x+4

We use the method of long division to get the answer. The method is shown in the first diagram below

$2x^{2}-8x+3$

Question 2:
$\frac{x}{6x-x^{2} } = \frac{x}{x(6-x)} = \frac{1}{6-x}$

Question 3:
$\frac{-12 x^{4} }{x^{4}+8 x^{5} }= \frac{-12 x^{4} }{ x^{4}(1+8x)}= \frac{-12}{1+8x}$

Question 4:
$\frac{x+5}{x^{2}+6x+5}= \frac{x+5}{(x+1)(x+5)}= \frac{1}{x+1}$

Question 5:
$\frac{x^{2}-3x-18} {x+3}= \frac{(x-6)(x+3)}{x+3}= \frac{x-6}{1}=x-6$

Question 6:
$\frac{2}{3a}$ × $\frac{2}{a^{2}}$ = $\frac{4}{3a^{3} }$ where $a \neq 0$

Question 7: (Question is not written well)
$\frac{x-5}{4x+8}$ × $(12x^{2}+32x+8)$
$\frac{12 x^{3}-28 x^{2} -152x-40 }{4x+8}$
By performing long division we get an answer $3 x^{2} -x-36$ with remainder of 248

Question 8:
$( \frac{x^{2}-16} {x-1})$ ÷ (x+4)

$( \frac{ x^{2}-16 }{x-1})$ × $\frac{1}{x+4}$
$\frac{(x+4)(x-1)}{x-1}$ × $\frac{1}{x+4}$
Cancelling out x+4

we obtain $\frac{x+1}{x-1}$

Question 9:
$\frac{x^{2}+2x+1} {x-2}$ ÷ $\frac{x^{2-1} }{x^{2}-4 }$
$\frac{ x^{2}+2x+1 }{x-2}$ × $\frac{x^{2}-4 }{x^{2}-1}$
Factorise all the quadratic expression gives
$\frac{(x+1)(x+1)}{x-2}$ × $\frac{(x-2)(x+2)}{(x+1)(x-1)}$
Cancelling out (x+1)

and

gives a simplest form
$\frac{(x+1)(x+2)}{x-1}$

Question 10:

$\frac{24 w^{10}+8w^{12} }{4 x^{4} }= \frac{24w^{10} }{4 x^{4} } + \frac{8 w^{12} }{4 x^{4} }$
Cancelling out the constants of each fraction
$\frac{6w^{10} }{x^{4} }+ \frac{2w^{12} }{x^{4}}= \frac{6w^{10}+2w^{12} }{ x^{4}}$

Question 11:

$\frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } = \frac{-2m^{6}(3m^{3}-3m^{2}-8)}{2m^{3} }$
Cancelling $2m^{3}$ gives us the simplified form
$-m^{3}(3m^{3}-3m^{2}-8) = -3m^{6}+3m^{5}+8m^{3}$

Question 12:

$\frac{-4x}{x+7} - \frac{8}{x-7} = \frac{-4x(x-7)-8(x+7)}{(x+7)(x-7)}$
$\frac{-4 x^{2} +28x-8x-56}{(x+7)(X-7)}= \frac{-4 x^{2} +20x-56}{(x+7)(x-7)}$
Factorising the numerator expression
$\frac{(-4x+28)(x-2)}{(x+7)(x-7)} = \frac{-4(x-7)(x-2)}{(x+7)(x-7)}$
Cancelling out x-7

gives the simplified form
$\frac{-4x+8}{x-7}$

Question 13:

$\frac{3}{x-3} - \frac{5}{x-2}= \frac{x3(x-2)-5(x-2)}{y(x-3)(x-2)}$
$\frac{3x-6-5x+15}{(x-3)(x-2)}= \frac{-2x+9}{(x-3)(x-2)}$

Question 14:

$\frac{9}{x-1}- \frac{5}{x+4}= \frac{9(x+4)-5(x-1)}{(x-1)(x+4)}$ $\frac{9x+36-5x+5}{(x-1)(x+4)}= \frac{4x+41}{(x-1)(x+4)}$

Question 15:

$\frac{-3}{x+2}- \frac{(-5)}{x+3}= \frac{-3(x+3)-(-5)(x+2)}{(x+2)(x+3)}$
$\frac{-3x-9+5x+10}{(x+2)(x+3)}= \frac{2x+1}{(x+2)(x+3)}$

Question 16:

$\frac{4}{x}+ \frac{5}{x}=-3$
$\frac{9}{x}=-3$
x=-3

Question 17:

$\frac{1}{3x-6}- \frac{5}{x-2}=12$
$\frac{(x-2)-5(3x-6)}{(3x-6)(x-2)} = \frac{x-2-15x+30}{(3x-6)(x-2)}= \frac{-14x+28}{(3x-6)(x-2)}$

Question 18

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is

Mathematics

Express the following in the form ax2+bx+c=0: (x+3)(3x - 2)=(4x +5)(2x-3) (3x+2)^2=(x+2)(x-3) (x+1)(x+2)=(2x-1)(x-2)

Step-by-step answer

P Answered by Specialist

9

Picture under is the answer for thisquestion
Express the following in the form ax2+bx+c=0: (x+3)(3x - 2)=(4x +5)(2x-3) (3x+2)^2=(x+2)(x-3) (x+1)