benjamindasilva20172
Answered

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Please help with these 2 questions and explain why it's the answer​

Please Help With These 2 Questions And Explain Why Its The Answer class=

Sagot :

belali

First, let's edit the given equations.

In 9:

[tex]2(32-x^2)=4x4-2\\\\64-2x^2=14[/tex]

Then, to check if [tex]x=5[/tex] is a solution to this equation, let's put the [tex]5[/tex].

[tex]64-2(5)^2=14\\\\64-50=14[/tex]

That's correct!

In 10:

[tex]4^2-x\leq x(6-5)\\\\16-x\leq 6x-5x\\\\16-x\leq x\\\\16\leq 2x\\\\x\geq 8[/tex]

Then, we must put [tex]15[/tex].

[tex]15\geq 8[/tex]

That's correct!

Good luck! If you have any question(s), then feel free to ask in comments!

igoroleshko156

Hi Student!

Let us first look at what the first question is asking for us to do and what is provided to do that.  Looking at the first question we see that they are asking if x = 5 is a solution to the expression that they provided.  This type of problem is fairly simple and what we basically do is just plug in the value 5 for the variable x in the expression and the simplify both sides to see if they are the same.

Plug in the values

  • [tex]2(32 - x^2) = 4 * 4 - 2[/tex]
  • [tex]2(32 - (5)^2) = 4 * 4 - 2[/tex]

Simplify the square and subtract inside of the parenthesis

  • [tex]2(32 - 25) = 4 * 4 - 2[/tex]
  • [tex]2(7) = 4 * 4 - 2[/tex]

Simplify both sides

  • [tex]2 * 7 = (4 * 4) - 2[/tex]
  • [tex]14 = (16) - 2[/tex]
  • [tex]14 = 14[/tex]

Looking at the final expression that we have, we can see that we have 14 is equal to 14 which is true and therefore x = 5 is a solution of the expression provided.

Moving onto the next question, we can see that it asks for us to do something really similar to the previous question but this time to check if x = 15 is a solution of a different expression.  Let's follow the same steps that we did last time.

Plug in the values

  • [tex]4^2 - x \le x(6 - 5)[/tex]
  • [tex]4^2 - (15) \le 15(6 - 5)[/tex]

Distribute the x

  • [tex]4^2 - (15) \le (15 * 6) + (15 * -5)[/tex]
  • [tex]4^2 - (15) \le 90 - 75[/tex]

Simplify both sides

  • [tex]16 - 15 \le 90 - 75[/tex]
  • [tex]1 \le 15[/tex]

Looking at the final expression that we have, we can see that it states that 1 is less than or equal to 15 and that is true.  Therefore, x = 15 is a solution of the expression that was provided.

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