Answer:
[tex]x>6[/tex]
Step-by-step explanation:
So we have the inequality:
[tex]-5\sqrt{x-2}<-10[/tex]
Divide both sides by -5. Since we're dividing by a negative, the sign changes:
[tex]\sqrt{x-2} >2[/tex]
Square both sides:
[tex]\sqrt{x-2}^2>2^2\\x-2>4[/tex]
Add two to both sides:
[tex](x-2)+2>4+2\\x>6[/tex]
Therefore, x>6.
However, the expression under the square root cannot be negative. Therefore, there is another inequality. Set the expression under the square root to greater than or equal to 0:
[tex]x-2\geq 0[/tex]
Add 2 to both sides:
[tex]x\geq 2[/tex]
However, previously, we determined that x must be greater than 6. That overrules this, so we will ignore it.
Thus, the answer is:
[tex]x>6[/tex]
So, drag the fourth option onto 6 on the line. This is an open circle and points to the right, so it includes all values greater than 6 not including 6.
