Respuesta :
Question 1
To find the width of the rectangle, we divide the area by the length
[tex]2x^{3}-29x+12 [/tex]÷[tex]x+4[/tex]
We use the method of long division to get the answer. The method is shown in the first diagram below
Answer: [tex]2x^{2}-8x+3[/tex]
Question 2:
[tex] \frac{x}{6x-x^{2} } = \frac{x}{x(6-x)} = \frac{1}{6-x} [/tex]
Question 3:
[tex] \frac{-12 x^{4} }{x^{4}+8 x^{5} }= \frac{-12 x^{4} }{ x^{4}(1+8x)}= \frac{-12}{1+8x} [/tex]
Question 4:
[tex] \frac{x+5}{x^{2}+6x+5}= \frac{x+5}{(x+1)(x+5)}= \frac{1}{x+1} [/tex]
Question 5:
[tex] \frac{x^{2}-3x-18} {x+3}= \frac{(x-6)(x+3)}{x+3}= \frac{x-6}{1}=x-6 [/tex]
Question 6:
[tex] \frac{2}{3a} [/tex]×[tex] \frac{2}{a^{2}} [/tex]=[tex] \frac{4}{3a^{3} } [/tex] where [tex]a \neq 0[/tex]
Question 7: (Question is not written well)
[tex] \frac{x-5}{4x+8} [/tex]×[tex](12x^{2}+32x+8) [/tex]
[tex] \frac{12 x^{3}-28 x^{2} -152x-40 }{4x+8} [/tex]
By performing long division we get an answer [tex]3 x^{2} -x-36[/tex] with remainder of 248
Question 8:
[tex]( \frac{x^{2}-16} {x-1}) [/tex]÷[tex](x+4)[/tex]
[tex]( \frac{ x^{2}-16 }{x-1}) [/tex]×[tex] \frac{1}{x+4} [/tex]
[tex] \frac{(x+4)(x-1)}{x-1} [/tex]×[tex] \frac{1}{x+4} [/tex]
Cancelling out [tex]x+4[/tex] we obtain [tex] \frac{x+1}{x-1} [/tex]
Question 9:
[tex] \frac{x^{2}+2x+1} {x-2} [/tex]÷[tex] \frac{x^{2-1} }{x^{2}-4 } [/tex]
[tex] \frac{ x^{2}+2x+1 }{x-2} [/tex]×[tex] \frac{x^{2}-4 }{x^{2}-1} [/tex]
Factorise all the quadratic expression gives
[tex] \frac{(x+1)(x+1)}{x-2} [/tex]×[tex] \frac{(x-2)(x+2)}{(x+1)(x-1)} [/tex]
Cancelling out [tex](x+1)[/tex] and [tex](x-2)[/tex] gives a simplest form
[tex] \frac{(x+1)(x+2)}{x-1} [/tex]
Question 10:
[tex] \frac{24 w^{10}+8w^{12} }{4 x^{4} }= \frac{24w^{10} }{4 x^{4} } + \frac{8 w^{12} }{4 x^{4} } [/tex]
Cancelling out the constants of each fraction
[tex] \frac{6w^{10} }{x^{4} }+ \frac{2w^{12} }{x^{4}}= \frac{6w^{10}+2w^{12} }{ x^{4}} [/tex]
Question 11:
[tex] \frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } = \frac{-2m^{6}(3m^{3}-3m^{2}-8)}{2m^{3} } [/tex]
Cancelling [tex]2m^{3} [/tex] gives us the simplified form
[tex] -m^{3}(3m^{3}-3m^{2}-8) = -3m^{6}+3m^{5}+8m^{3} [/tex]
Question 12:
[tex] \frac{-4x}{x+7} - \frac{8}{x-7} = \frac{-4x(x-7)-8(x+7)}{(x+7)(x-7)} [/tex]
[tex] \frac{-4 x^{2} +28x-8x-56}{(x+7)(X-7)}= \frac{-4 x^{2} +20x-56}{(x+7)(x-7)} [/tex]
Factorising the numerator expression
[tex] \frac{(-4x+28)(x-2)}{(x+7)(x-7)} = \frac{-4(x-7)(x-2)}{(x+7)(x-7)} [/tex]
Cancelling out [tex]x-7[/tex] gives the simplified form
[tex] \frac{-4x+8}{x-7} [/tex]
Question 13:
[tex] \frac{3}{x-3} - \frac{5}{x-2}= \frac{x3(x-2)-5(x-2)}{y(x-3)(x-2)} [/tex]
[tex] \frac{3x-6-5x+15}{(x-3)(x-2)}= \frac{-2x+9}{(x-3)(x-2)} [/tex]
Question 14:
[tex] \frac{9}{x-1}- \frac{5}{x+4}= \frac{9(x+4)-5(x-1)}{(x-1)(x+4)} [/tex][tex] \frac{9x+36-5x+5}{(x-1)(x+4)}= \frac{4x+41}{(x-1)(x+4)} [/tex]
Question 15:
[tex] \frac{-3}{x+2}- \frac{(-5)}{x+3}= \frac{-3(x+3)-(-5)(x+2)}{(x+2)(x+3)} [/tex]
[tex] \frac{-3x-9+5x+10}{(x+2)(x+3)}= \frac{2x+1}{(x+2)(x+3)} [/tex]
Question 16:
[tex] \frac{4}{x}+ \frac{5}{x}=-3 [/tex]
[tex] \frac{9}{x}=-3 [/tex]
[tex]x=-3[/tex]
Question 17:
[tex] \frac{1}{3x-6}- \frac{5}{x-2}=12 [/tex]
[tex] \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)} [/tex]
Question 18
To find the width of the rectangle, we divide the area by the length
[tex]2x^{3}-29x+12 [/tex]÷[tex]x+4[/tex]
We use the method of long division to get the answer. The method is shown in the first diagram below
Answer: [tex]2x^{2}-8x+3[/tex]
Question 2:
[tex] \frac{x}{6x-x^{2} } = \frac{x}{x(6-x)} = \frac{1}{6-x} [/tex]
Question 3:
[tex] \frac{-12 x^{4} }{x^{4}+8 x^{5} }= \frac{-12 x^{4} }{ x^{4}(1+8x)}= \frac{-12}{1+8x} [/tex]
Question 4:
[tex] \frac{x+5}{x^{2}+6x+5}= \frac{x+5}{(x+1)(x+5)}= \frac{1}{x+1} [/tex]
Question 5:
[tex] \frac{x^{2}-3x-18} {x+3}= \frac{(x-6)(x+3)}{x+3}= \frac{x-6}{1}=x-6 [/tex]
Question 6:
[tex] \frac{2}{3a} [/tex]×[tex] \frac{2}{a^{2}} [/tex]=[tex] \frac{4}{3a^{3} } [/tex] where [tex]a \neq 0[/tex]
Question 7: (Question is not written well)
[tex] \frac{x-5}{4x+8} [/tex]×[tex](12x^{2}+32x+8) [/tex]
[tex] \frac{12 x^{3}-28 x^{2} -152x-40 }{4x+8} [/tex]
By performing long division we get an answer [tex]3 x^{2} -x-36[/tex] with remainder of 248
Question 8:
[tex]( \frac{x^{2}-16} {x-1}) [/tex]÷[tex](x+4)[/tex]
[tex]( \frac{ x^{2}-16 }{x-1}) [/tex]×[tex] \frac{1}{x+4} [/tex]
[tex] \frac{(x+4)(x-1)}{x-1} [/tex]×[tex] \frac{1}{x+4} [/tex]
Cancelling out [tex]x+4[/tex] we obtain [tex] \frac{x+1}{x-1} [/tex]
Question 9:
[tex] \frac{x^{2}+2x+1} {x-2} [/tex]÷[tex] \frac{x^{2-1} }{x^{2}-4 } [/tex]
[tex] \frac{ x^{2}+2x+1 }{x-2} [/tex]×[tex] \frac{x^{2}-4 }{x^{2}-1} [/tex]
Factorise all the quadratic expression gives
[tex] \frac{(x+1)(x+1)}{x-2} [/tex]×[tex] \frac{(x-2)(x+2)}{(x+1)(x-1)} [/tex]
Cancelling out [tex](x+1)[/tex] and [tex](x-2)[/tex] gives a simplest form
[tex] \frac{(x+1)(x+2)}{x-1} [/tex]
Question 10:
[tex] \frac{24 w^{10}+8w^{12} }{4 x^{4} }= \frac{24w^{10} }{4 x^{4} } + \frac{8 w^{12} }{4 x^{4} } [/tex]
Cancelling out the constants of each fraction
[tex] \frac{6w^{10} }{x^{4} }+ \frac{2w^{12} }{x^{4}}= \frac{6w^{10}+2w^{12} }{ x^{4}} [/tex]
Question 11:
[tex] \frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } = \frac{-2m^{6}(3m^{3}-3m^{2}-8)}{2m^{3} } [/tex]
Cancelling [tex]2m^{3} [/tex] gives us the simplified form
[tex] -m^{3}(3m^{3}-3m^{2}-8) = -3m^{6}+3m^{5}+8m^{3} [/tex]
Question 12:
[tex] \frac{-4x}{x+7} - \frac{8}{x-7} = \frac{-4x(x-7)-8(x+7)}{(x+7)(x-7)} [/tex]
[tex] \frac{-4 x^{2} +28x-8x-56}{(x+7)(X-7)}= \frac{-4 x^{2} +20x-56}{(x+7)(x-7)} [/tex]
Factorising the numerator expression
[tex] \frac{(-4x+28)(x-2)}{(x+7)(x-7)} = \frac{-4(x-7)(x-2)}{(x+7)(x-7)} [/tex]
Cancelling out [tex]x-7[/tex] gives the simplified form
[tex] \frac{-4x+8}{x-7} [/tex]
Question 13:
[tex] \frac{3}{x-3} - \frac{5}{x-2}= \frac{x3(x-2)-5(x-2)}{y(x-3)(x-2)} [/tex]
[tex] \frac{3x-6-5x+15}{(x-3)(x-2)}= \frac{-2x+9}{(x-3)(x-2)} [/tex]
Question 14:
[tex] \frac{9}{x-1}- \frac{5}{x+4}= \frac{9(x+4)-5(x-1)}{(x-1)(x+4)} [/tex][tex] \frac{9x+36-5x+5}{(x-1)(x+4)}= \frac{4x+41}{(x-1)(x+4)} [/tex]
Question 15:
[tex] \frac{-3}{x+2}- \frac{(-5)}{x+3}= \frac{-3(x+3)-(-5)(x+2)}{(x+2)(x+3)} [/tex]
[tex] \frac{-3x-9+5x+10}{(x+2)(x+3)}= \frac{2x+1}{(x+2)(x+3)} [/tex]
Question 16:
[tex] \frac{4}{x}+ \frac{5}{x}=-3 [/tex]
[tex] \frac{9}{x}=-3 [/tex]
[tex]x=-3[/tex]
Question 17:
[tex] \frac{1}{3x-6}- \frac{5}{x-2}=12 [/tex]
[tex] \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)} [/tex]
Question 18
** CORRECTIONS: Q1: It's 2x^3-29x+12; Q2,3,4,5,6: All conditions have ≠ symbol; Q7: it's (12x^2+32x+16); Q10: Option D should be divided by x^4; **
(1) Given:
Width = W = x+4
Area = A = [tex] 2x^3-29x+12 [/tex]
Length = L = ?
Since the pool is rectangular in shape:
area = width * length
A = W * L
Substitute:
[tex] 2x^3-29x+12 = (x+4) * L \\ L =\frac{2x^3-29x+12}{x+4} [/tex]
The long division is attached with the answer (below in the picture). Hence the correct answer is [tex] 2x^2-8x+3 [/tex] (Option C)
(2) Given expression:
[tex] \frac{x}{6x-x^2} \\ \frac{x}{x(6-x)} \\ \frac{1}{6-x} [/tex]
Where x ≠ 6. (Option B)
(3) Given :
[tex] \frac{-12 x^{4} }{x^{4}+8 x^{5} } [/tex]
Now simplify:
[tex] \frac{-12 x^{4} }{x^{4}+8 x^{5} }= \frac{-12 x^{4} }{ x^{4}(1+8x)}= \frac{-12}{1+8x} [/tex]
Where x ≠ -1/8 (Option A)
(4) Given:
[tex] \frac{x+5}{x^{2}+6x+5}[/tex]
Simplify:
[tex] \frac{x+5}{x^{2}+6x+5}= \frac{x+5}{(x+1)(x+5)}= \frac{1}{x+1} [/tex]
Where x ≠ -1 (Option A)
(5) Given:
[tex] \frac{x^{2}-3x-18} {x+3} [/tex]
Simplify:
[tex] \frac{x^{2}-3x-18} {x+3}= \frac{(x-6)(x+3)}{x+3}= \frac{x-6}{1}=x-6 [/tex] where x≠6 (Option C)
(6) Given:
[tex] \frac{2}{3a} .\frac{2}{a^2} [/tex]
Simplify:
[tex] \frac{4}{3a^{1+2}} = \frac{4}{3a^{3}} [/tex]
Where a ≠ 0 (Option C)
(7) Mathematically:
[tex] \frac{x-5}{4x + 8} * (12x^2+32x+16) [/tex]
Simplify:
[tex] \frac{x-5}{4x + 8} * (12x^2+32x+16) \\ \frac{x-5}{4(x + 2)} * 12x^2 + \frac{x-5}{4(x + 2)} * 32x + \frac{x-5}{4(x + 2)} * 16 \\ \frac{x-5}{(x + 2)} * 3x^2 + \frac{x-5}{(x + 2)} * 8x + \frac{x-5}{(x + 2)} * 4 \\ \frac{(x-5)(3x^2 + 8x + 4x) }{(x+2)} \\ \frac{(x-5)(3x^2 -6x - 2x + 4x) }{(x+2)} \\ \frac{(x-5)(3x+2)(x+2) }{(x+2)} \\ =(x-5)(3x+2) [/tex]
(Option C)
(8) Simplify:
[tex] \frac{( \frac{x^{2}-16} {x-1}) }{(x+4)} \\ \frac{( \frac{(x+4)(x-4)} {x-1}) }{(x+4)} \\ = \frac{(x-4)} {(x-1)} [/tex]
(Option A)
(9) Simplify:
[tex] \frac{ \frac{x^2+2x+1}{x-2}}{\frac{x^2-1}{x^2-4 }} \\ \frac{ \frac{(x+1)(x+1)}{x-2}}{\frac{(x+1)(x-1)}{(x-2)(x+2) }} \\ \frac{ \frac{(x+1)}{1}}{\frac{(x-1)}{(x+2) }} \\ = \frac{(x+1)(x+2)}{(x-1)} [/tex]
(Option A)
(10) Given:
[tex] \frac{24 w^{10}+8w^{12} }{4 x^{4} } [/tex]
Simplify:
[tex] \frac{24 w^{10}+8w^{12} }{4 x^{4} }= \frac{24w^{10} }{4 x^{4} } + \frac{8 w^{12} }{4 x^{4} } = \frac{6w^{10} }{x^{4} }+ \frac{2w^{12} }{x^{4}}= \frac{6w^{10}+2w^{12} }{ x^{4}} [/tex]
(Option D)
(11) Given:
[tex] \frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } [/tex]
Simplify:
[tex] \frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } = \frac{-2m^{6}(3m^{3}+3m^{2}+8)}{2m^{3} } = -m^{3}(3m^{3}+3m^{2}+8)\\ = -3m^{6}-3m^{5}-8m^{3} [/tex]
(Option C)
(12) Simplify:
[tex] \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)} \\ \frac{(-4x+28)(x-2)}{(x+7)(x-7)} = \frac{-4(x-7)(x-2)}{(x+7)(x-7)} = \frac{-4x+8}{x+7} [/tex]
(Option A)
(13) Simplify:
[tex] \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)} [/tex]
(Option A)
(14) Simplify:
[tex] \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)} [/tex]
(Option B)
(15) Simplify:
[tex] \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)} [/tex]
(Option D)
(16) Given:
4/x + 5/x = -3
Simplify:
(4+5)/x = -3
-3x = 9
x = -3 (Option C)
(17) Simplify:
[tex] \frac{1}{3x-6} - \frac{5}{x-2} = 12 \\ \frac{(x-2)-5(3x-6)}{(3x-6)(x-2)} = 12 \\ \frac{(x-2)-5*3(x-2)}{(3x-6)(x-2)} = 12 \\ \frac{-14(x-2)}{(3x-6)(x-2)} = 12 \\ \frac{-14}{(3x-6)} = 12\\ -14 = 12(3x-6) \\ -14 = 36x - 72 \\ 36x = 58 \\ x=\frac{29}{18} [/tex]
(Option D)
(18) Simplify:
[tex] \frac{1}{x} - \frac{6}{x^2} = -12 \\ \frac{x - 6}{x^2} = -12 \\ x-6 = -12x^2 \\ 12x^2 + x - 6 = 0 \\ 12x^2 + 9x - 8x - 6 = 0 \\ 3x(4x + 3) -2(4x + 3) =0 \\ (3x-2)(4x+3) =0 \\ => x =\frac{2}{3} , x =\frac{-3}{4} [/tex]
(Option C)
(19) Dorothy's rate (alone) will be:
[tex] R_D =\frac{1}{6} [/tex]
Rosanne's rate (alone) will be:
[tex] R_R =\frac{1}{8} [/tex]
If both work together, add both the rates:
[tex] R_T = R_D + R_R = \frac{1}{6} + \frac{1}{8} = \frac{7}{24} [/tex] (in 1/hours)
To find the hours, flip the rate:
[tex] \frac{24}{7} = 3.43 [/tex] hours (Option B)
(20) As pressure (p) is inversely proportional with volume (v):
p = k/v (where k is constant of proportionality)
k = pv
Find constant using initial values:
k = (104)(108)
k = 11232
Now new pressure is:
p = k/v = 11232/432 = 26 Pa (Option A)
(21)
x: 1,3,5,10
y: 4,12,20,40
Direct variation is the value of y increases with x. So,
y = 4x
If x = 1,y=4(1)=4
If x = 3,y=4(3)=12
If x = 5,y=20
If x = 10,y=40 (Option A)
(22) [tex] \frac{3}{4x+64} [/tex]
If x=-16,4(-16) + 64 = 0;denominator will become zero,which means that there will be discontinuity at x = -16. Hence, x=-16 (Option C) should be excluded.
