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Mathematics
, 25.07.2019 09:00,
vondah4014
What is the solution to the equation
Answers: 1
Show answers
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answered:
catty7343
The solution of your problem is shown on the picture below.
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answered: Guest
-14 is the answer
hope this : ))
answered: Guest
its not b
step-by-step explanation:
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Mathematics, 23.06.2019 07:20,
ANONYMUSNESS8670
Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution. u = x 4(x +1) x +1 (x + 1)2 select the solution(s) of the original equation. x = 1 + sqrt 2 x = 2 + sqrt 2 x = 3 + sqrt 2 x = 1 - sqrt 2 x = 2 - sqrt 2 x = 3 - sqrt 2
Answers: 1
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Mathematics, 14.07.2019 19:00,
laurenppylant
Solve the equation and express each solution in a+bi form. x^4-7x^2-8=0 a. x = -1, 1, 2 sqrt 2, or -2 sqrt 2 b. x = -i, i, 2 sqrt 2, or -2 sqrt 2 c. x = -1, 1, 2 sqrt 2i, or -2 sqrt 2i d. x = -i, i, 2 sqrt 2i, or -2 sqrt 2i
Answers: 1
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Mathematics, 26.09.2019 05:30,
5001Je
2. the equation x+2= sqrt (3x+10) is of the form x+a=sqrt(bx+c) , where a, b, and c are all positive integers and b > 1. using this equation as a model, create your own equation that has extraneous solutions. (a) using trial and error with numbers for a, b, and c, create an equation of the form x+a=sqrt(bx+c) where a, b, and c are all positive integers and b > 1 such that 7 is a solution. (hint: substitute 7 for x, and choose a value for a. then square both sides so you can choose a, b, and c that will make the equation true.) (b) solve the equation you created in part 2a. (c) if your solution in part 2b did not have an extraneous solution, revise your equation so that 7 is one solution and there is an extraneous solution. if your solution in part 2b did have an extraneous solution, create another equation with different values of a, b, and c that also has 7 as one solution and an extraneous solution.
Answers: 1
continue
Mathematics, 25.10.2019 18:43,
Jana1517
The equation x+2=sqrt(3x+10) is of the form , x+a=sqrt(bx+c),where a, b, and c are all positive integers and b> 1 . using this equation as a model, create your own equation that has extraneous solutions. using trial and error with numbers for a, b, and c, create an equation of the form x+a=sqrt(bx+c) , where a, b, and c are all positive integers and b> 1 such that 7 is a solution. (hint: substitute 7 for x, and choose a value for a. then square both sides so you can choose a, b, and c that will make the equation true.) solve the equation you created in part 2a. if your solution in part 2b did not have an extraneous solution, revise your equation so that 7 is one solution and there is an extraneous solution. if your solution in part 2b did have an extraneous solution, create another equation with different values of a, b, and c that also has 7 as one solution and an extraneous solution.
Answers: 1
continue
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What is the solution to the equation [tex] - \sqrt{x + 10} = 7[/tex]...
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