|4x +2|=10 confirm your solution using a graph or table

Answers

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

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

TheOriginal2x

Oh yikes this is a scary question

Step-by-step explanation:

Good luck bro!

honey66

it is c

hope i helped

Step-by-step explanation:

maya8909

carnival one f(x) = 0.5x - money per ticket

g(x) = 8x - tickets per hour

We replace g(x) with its value (8x).

f(g(x)) = f(8x)

We replace the x in f(x) with 8x.

f(8x) = 0,5 × 8x

f(8x) = 4x

⇒ f(g(x)) = 4x

Correct

a. f(g(x)) = 4x, which represents the money Aurora made in dollars per hour

Step-by-step explanation:

leandroarguijo

answer:

Step-by-step explanation:

enicole10

1. 4x⁶ + 6x⁵ + 8x⁴ is your answer, so b.

2. x^2 − 49 is your answer, so b.

3. 2x^2 + 6x-20 is your answer, so a.

4. (f • g)(2) = 12, the distance in miles the bicycle traveled, so b. ( not 100% sure tho)

5. 8x³-34x²+25x-3 is your answer, so c.

6. 30x^2 - 26x -12 is polynomial is the answer, so c.

7. not sure : (

8. not sure: (

9. not sure : (

10. x^2 − 14x + 49 is the answer, so a.

hope this ! : )

Tracic1972

1) 10x⁵+38x⁴-23x³-101x²- 96x+50

-x+17

(7x²-x-29)=0

3x³ -13x²-4x + 67=0

3x³ -6x²-11x+19

5x²- 20x

Step-by-step explanation:

Spanish

Agregar los polinomios:

(4x + 9) + (5x - 8) (x2 - 4x + 7) + (2x2 + 6x - 8) (5x3 + 4x2 - 7x -8) + (2x3 - 3x2 + 4x - 3)

Primero multiplicaremos los polinomios con los corchetes

= 4x + 9 + (5x³- 20x² + 35x - 16x² + 32x-56) + (10x⁵ + 8x⁴- 14x³- 16x² + 30x⁴ + 24x³- 42x²- 48x- 40x³- 32x²- 56x + 64) + (2x³- 42x² + 4x -3)

Ahora simplificaremos dentro de los corchetes agregando primero los términos similares

= 4x + 9 + (5x³- 36x² + 67x-56) + (10x⁵ + 38x⁴- 30x³- 90x²-104x + 64) + (2x³- 42x² + 4x-3)

Ahora suma o resta todos los términos similares y distintos

= 10x⁵ + 38x⁴- 30x³ + 5x³ + 2x³- 90x²- 36x²- 42x² + 67x²-104x + 4x + 4x + 64 + 9-3

= 10x⁵ + 38x⁴-23x³-101x²- 96x + 50 Ans

Restando los polinomios

(4x + 9) - (5x - 8) = - x + 17

-5x +8 + 4x + 9 = -x + 17

-x + 17 = -x + 17

(x2 - 4x + 7) - (2x2 + 6x - 8) = -20x2-13x-72

x²-2x²-4x-6x + 7 + 8 = 20x2-13x-72

-x²-10x + 15-20x² + 13x + 72 = 0

-21x² + 3x + 87 = 0

-3 (7x²-x-29) = 0

(7x²-x-29) = 0

(5x3 + 4x2 - 7x -8) - (2x3 - 3x2 + 4x - 3) = 20x2 + 32x-45x-72

5x³-2x³ + 4x² + 3x²-7x-4x-8 + 3 = 20x2 -7x-72

3x³ + 7x²-11x-5-20x² + 7x + 72 = 0

3x³ -13x²-4x + 67 = 0

(7x3 - 10x2 - 11x + 9) - (4x3 - 4x2 - 10)

Los signos negativos cambian cuando restamos

= 3x³ -6x²-11x + 19

Ahora multiplicando

5x (x-4) = 5x²- 20x

English

Adding the polynomials :

(4x + 9) + (5x – 8) (x2 – 4x + 7) + (2x2 + 6x – 8) (5x3 + 4x2 – 7x -8) + (2x3 – 3x2 + 4x – 3)

First we will multiply the polynomials with the brackets

= 4x+9 + ( 5x³- 20x²+ 35x - 16x²+ 32x-56) + (10x⁵ + 8x⁴- 14x³- 16x²+30x⁴+ 24x³- 42x²- 48x- 40x³- 32x²- 56x+ 64) + ( 2x³- 42x²+ 4x-3)

Now we will simplify within the brackets first adding the like terms

= 4x+9 + (5x³- 36x²+67x-56) + (10x⁵ +38x⁴- 30x³- 90x²-104x+64) + ( 2x³- 42x²+ 4x-3)

Now add or subtract all the like and unlike terms

= 10x⁵+38x⁴- 30x³+5x³+2x³- 90x²- 36x²- 42x²+67x²-104x+4x+4x+64+9-3

= 10x⁵+38x⁴-23x³-101x²- 96x+50 Ans

Subtracting the Polynomials

(4x + 9) - (5x – 8)=-x+17

-5x +8 +4x+9 = -x+17

-x+17=-x+17

(x2 – 4x + 7) - (2x2 + 6x – 8)= -20x2-13x-72

x²-2x²-4x-6x+7+8=20x2-13x-72

-x²-10x+ 15-20x²+13x+72=0

-21x²+3x+87=0

-3(7x²-x-29)=0

(7x²-x-29)=0

(5x3 + 4x2 – 7x -8) - (2x3 – 3x2 + 4x – 3)= 20x2 +32x-45x-72

5x³-2x³+4x²+3x²-7x-4x-8+3= 20x2 -7x-72

3x³+7x²-11x-5-20x²+7x+72=0

3x³ -13x²-4x + 67=0

(7x3 – 10x2 – 11x + 9) - (4x3 – 4x2 – 10)

The negative signs are changed when we subtract

= 3x³ -6x²-11x+19

Now Multiplying

5x(x-4) = 5x²- 20x

javink18

** 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 = 2x^3-29x+12

Length = L = ?

Since the pool is rectangular in shape:

area = width * length

A = W * L

Substitute:

$2x^3-29x+12 = (x+4) * L \\ L =\frac{2x^3-29x+12}{x+4}$

The long division is attached with the answer (below in the picture). Hence the correct answer is 2x^2-8x+3 (Option C)

(2) Given expression:

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

Where x ≠ 6. (Option B)

(3) Given :

$\frac{-12 x^{4} }{x^{4}+8 x^{5} }$

Now simplify:

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

Where x ≠ -1/8 (Option A)

(4) Given:

$\frac{x+5}{x^{2}+6x+5}$

Simplify:

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

Where x ≠ -1 (Option A)

(5) Given:

$\frac{x^{2}-3x-18} {x+3}$

Simplify:

$\frac{x^{2}-3x-18} {x+3}= \frac{(x-6)(x+3)}{x+3}= \frac{x-6}{1}=x-6$ where x≠6 (Option C)

(6) Given:

$\frac{2}{3a} .\frac{2}{a^2}$

Simplify:

$\frac{4}{3a^{1+2}} = \frac{4}{3a^{3}}$

Where a ≠ 0 (Option C)

(7) Mathematically:

$\frac{x-5}{4x + 8} * (12x^2+32x+16)$

Simplify:

$\frac{x-5}{4x + 8} * (12x^2+32x+16) \\ \frac{x-5}{4(x + 2)} * 12x^2 + \frac{x-5}{4(x + 2)} * 32x + \frac{x-5}{4(x + 2)} * 16 \\ \frac{x-5}{(x + 2)} * 3x^2 + \frac{x-5}{(x + 2)} * 8x + \frac{x-5}{(x + 2)} * 4 \\ \frac{(x-5)(3x^2 + 8x + 4x) }{(x+2)} \\ \frac{(x-5)(3x^2 -6x - 2x + 4x) }{(x+2)} \\ \frac{(x-5)(3x+2)(x+2) }{(x+2)} \\ =(x-5)(3x+2)$

(Option C)

(8) Simplify:

$\frac{( \frac{x^{2}-16} {x-1}) }{(x+4)} \\ \frac{( \frac{(x+4)(x-4)} {x-1}) }{(x+4)} \\ = \frac{(x-4)} {(x-1)}$

(Option A)

(9) Simplify:

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

(Option A)

(10) Given:

$\frac{24 w^{10}+8w^{12} }{4 x^{4} }$

Simplify:

$\frac{24 w^{10}+8w^{12} }{4 x^{4} }= \frac{24w^{10} }{4 x^{4} } + \frac{8 w^{12} }{4 x^{4} } = \frac{6w^{10} }{x^{4} }+ \frac{2w^{12} }{x^{4}}= \frac{6w^{10}+2w^{12} }{ x^{4}}$

(Option D)

(11) Given:

$\frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} }$

Simplify:

$\frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } = \frac{-2m^{6}(3m^{3}+3m^{2}+8)}{2m^{3} } = -m^{3}(3m^{3}+3m^{2}+8)\\ = -3m^{6}-3m^{5}-8m^{3}$

(Option C)

(12) Simplify:

$\frac{-4x}{x+7} - \frac{8}{x-7} = \frac{-4x(x-7)-8(x+7)}{(x+7)(x-7)} \\ \frac{-4 x^{2} +28x-8x-56}{(x+7)(X-7)}= \frac{-4 x^{2} +20x-56}{(x+7)(x-7)} \\ \frac{(-4x+28)(x-2)}{(x+7)(x-7)} = \frac{-4(x-7)(x-2)}{(x+7)(x-7)} = \frac{-4x+8}{x+7}$

(Option A)

(13) Simplify:

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

(Option A)

(14) Simplify:

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

(Option B)

(15) Simplify:

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

(Option D)

(16) Given:

4/x + 5/x = -3

Simplify:

(4+5)/x = -3

-3x = 9

x = -3 (Option C)

(17) Simplify:

$\frac{1}{3x-6} - \frac{5}{x-2} = 12 \\ \frac{(x-2)-5(3x-6)}{(3x-6)(x-2)} = 12 \\ \frac{(x-2)-5*3(x-2)}{(3x-6)(x-2)} = 12 \\ \frac{-14(x-2)}{(3x-6)(x-2)} = 12 \\ \frac{-14}{(3x-6)} = 12\\ -14 = 12(3x-6) \\ -14 = 36x - 72 \\ 36x = 58 \\ x=\frac{29}{18}$

(Option D)

(18) Simplify:

$\frac{1}{x} - \frac{6}{x^2} = -12 \\ \frac{x - 6}{x^2} = -12 \\ x-6 = -12x^2 \\ 12x^2 + x - 6 = 0 \\ 12x^2 + 9x - 8x - 6 = 0 \\ 3x(4x + 3) -2(4x + 3) =0 \\ (3x-2)(4x+3) =0 \\ = x =\frac{2}{3} , x =\frac{-3}{4}$

(Option C)

(19) Dorothy's rate (alone) will be:

$R_D =\frac{1}{6}$

Rosanne's rate (alone) will be:

$R_R =\frac{1}{8}$

If both work together, add both the rates:

$R_T = R_D + R_R = \frac{1}{6} + \frac{1}{8} = \frac{7}{24}$ (in 1/hours)

To find the hours, flip the rate:

$\frac{24}{7} = 3.43$ 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) $\frac{3}{4x+64}$

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.

jayv1999

x = -3 or +2

Step-by-step explanation:

Such an equation resolves into two equations. The argument of the absolute value function can be either positive or negative, so there is one equation for each case.

4x +2 = 10

4x = 8 . . . . subtract 2

x = 2 . . . . . divide by 4

4x +2 = -10

4x = -12 . . . subtract 2

x = -3 . . . . . divide by 4

The two solutions are x=-3 and x=2.

|4x +2|=10 confirm your solution using a graph or table

Jazzy2006

Remove the absolute value term. 4x+2=10 solve this, x=2

Set up the negative portion 4x+2= -10 solve this, x= -3

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