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1. What is the solution to the equation?

2.Let f(x)=−2x+4 and g(x)=−6−7. Find f(x)−g(x)

1. What is the solution to the equation? 2.Let f(x)=−2x+4 and g(x)=−6−7. Find f(x-example-1
1. What is the solution to the equation? 2.Let f(x)=−2x+4 and g(x)=−6−7. Find f(x-example-1
1. What is the solution to the equation? 2.Let f(x)=−2x+4 and g(x)=−6−7. Find f(x-example-2

2 Answers

6 votes

Answer:


Explanation:

Problem One

  • sqrt(2x + 10) - 6 = 2 Add 6 to both sides
  • sqrt(2x + 10) = 2 + 6
  • sqrt(2x + 10) = 8 Square both sides.
  • sqrt(2x + 10)^2 = 8^2 Do the operation
  • 2x + 10 = 64 Subtract 10 from both sides
  • 2x + 10 - 10 = 64 - 10 Combine
  • 2x = 54 Divide by 2
  • 2x/2 = 54/2
  • x = 27

Problem Two

Problem two has a small problem. The way it is written suggests that the question number might be seven. Maybe that's why you get a second answer.

f(x) = - 2x + 4

g(x) = -6 - 7 I'm assuming the 7 is a counting number and not the question number

g(x) = - 13

f(x) - g(x) = -2x + 4 - - 13

f(x) - g(x) = -2x + 4 + 13

f(x) - g(x) = - 2x + 17

Now suppose it is the question number

f(x) - g(x) = - 2X + 4 - 6

f(x) - g(x) = - 2x - 2

User Ajay Tom George
by
5.6k points
2 votes

Answer:

  1. x = 27
  2. (f-g)(x) = 17-2x

Explanation:

1.


√(2x+10)-6=2\\\\√(2x+10)=8 \qquad\text{add 6}\\\\2x+10=64 \qquad\text{square both sides}\\\\2x=54 \qquad\text{subtract 10}\\\\x=27 \qquad\text{divide by 2}

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2. f(x) -g(x) = (-2x +4) -(-6 -7) = -2x +4 +6 +7

f(x) -g(x) = 17 -2x