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frez [133]
3 years ago
8

(5y + 2) + 14x = 5y + (2 + 14x)

Mathematics
2 answers:
Ann [662]3 years ago
5 0
6. (5y+2)+14x=5y+(2+14x) (1point) A.associative property of addition B.associative property of multiplication C.commutative property of addition D.commutative property of multiplication 7. (m*n)*p=m*(n*p) (1 point) A. commutative property of addition B.commutative property of multiplication C.associative property of addition D. associative property of multiplication
Lorico [155]3 years ago
3 0
Simplify brackets

<span>5y+2+14x=5y+2+14x<span>5y+2+14x=5y+2+14x</span></span>
Since both sides equal, there are infinitely many solutions
Infinitely Many Solutions

SO BASICALLY THAT MEANS IT EQUALS TO 0. 

<span>Simplifying (5y + 2) + 14x = 5y + (2 + 14x) Reorder the terms: (2 + 5y) + 14x = 5y + (2 + 14x) Remove parenthesis around (2 + 5y) 2 + 5y + 14x = 5y + (2 + 14x) Reorder the terms: 2 + 14x + 5y = 5y + (2 + 14x) Remove parenthesis around (2 + 14x) 2 + 14x + 5y = 5y + 2 + 14x Reorder the terms: 2 + 14x + 5y = 2 + 14x + 5y Add '-2' to each side of the equation. 2 + 14x + -2 + 5y = 2 + 14x + -2 + 5y Reorder the terms: 2 + -2 + 14x + 5y = 2 + 14x + -2 + 5y Combine like terms: 2 + -2 = 0 0 + 14x + 5y = 2 + 14x + -2 + 5y 14x + 5y = 2 + 14x + -2 + 5y Reorder the terms: 14x + 5y = 2 + -2 + 14x + 5y Combine like terms: 2 + -2 = 0 14x + 5y = 0 + 14x + 5y 14x + 5y = 14x + 5y Add '-14x' to each side of the equation. 14x + -14x + 5y = 14x + -14x + 5y Combine like terms: 14x + -14x = 0 0 + 5y = 14x + -14x + 5y 5y = 14x + -14x + 5y Combine like terms: 14x + -14x = 0 5y = 0 + 5y 5y = 5y Add '-5y' to each side of the equation. 5y + -5y = 5y + -5y Combine like terms: 5y + -5y = 0 0 = 5y + -5y Combine like terms: 5y + -5y = 0 0 = 0 Solving 0 = 0 Couldn't find a variable to solve for.</span>
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An equation is given. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places w
Rina8888 [55]

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a) The general solution   θ = nπ±π/6

The all solutions are  - π/6 ,π/6 ,5π/6 ,7π/6 ,11π/6 ,13π/6 ...…

b)   The solutions in the interval [ 0,2π)

                            θ =  - π/6 ,π/6 , 5π/6 , 7π/6 , 11π/6

Step-by-step explanation:

<u><em>Step( i)</em></u>:-

Given an equation 2 cos (2θ)-1 =0

                               2 cos (2θ) = 1

                                cos(2θ) = 1/2

                                cos(2θ) = cos (π/3)

<em>Step(ii):-</em>

<em>a) </em>

The general solution of  cos θ = cos ∝ is given by

                                              θ = 2nπ±∝

The general solution of  cos(2θ) = cos (π/3) is

                                               2θ = 2nπ±π/3

                                                θ = nπ±π/6

Put n=0          θ =  ±π/6

Put n =1          θ = π±π/6

                       θ = π-π/6 = 5π/6

                        θ = π+π/6 = 7π/6

put n =2

                        θ = 2π±π/6  

                         θ = 2π-π/6 = 11π/6

                        θ = 2π+π/6 = 13π/6

   

The all solutions are  - π/6 ,π/6 ,5π/6 ,7π/6 ,11π/6 ,13π/6 ...…

b)

        The solutions in the interval [ 0,2π)

                            θ =  - π/6 ,π/6 , 5π/6 , 7π/6 , 11π/6

<u><em>Final answer</em></u>:-

a)

The all solutions are  - π/6 ,π/6 ,5π/6 ,7π/6 ,11π/6 ,13π/6 ...…

b)

The solutions in the interval [ 0,2π)

                            θ =  - π/6 ,π/6 , 5π/6 , 7π/6 , 11π/6

             

                                               

                                         

8 0
3 years ago
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