The difference of six and a number is greater then nine

Mathematics

1 answer:

Dmitry [639]3 years ago

8 0

6-m>9
difference would be subtraction of 6, 6 would go first then a random letter and greater than > 9.

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Solve for x.<br> Sqrt8 (Sqrt2 - x) = 11

Alexus [3.1K]

Answer: $\frac{-7\sqrt{2} }{4}$

Step-by-step explanation:

So first, let's get rid of the parentheses. We can multiply it out to get $\sqrt{16}-x\sqrt{8} =11$ . We know that the square root of 16 is ±4, so now our equation is ±4 - $x\sqrt{8}$ =11. I'm guessing since the problem only has one solution it's most likely only positive 4, so let's revise our equation to 4 - $x\sqrt{8}$ =11. We can use inverse operations to make the a little easier to solve: -7= $x\sqrt{8}$ . We divide both sides by $\sqrt{8}$ to get $\frac{-7}{\sqrt{8} } =x$ , which we can rationalize (remove the square root from the denominator so that it's a proper answer) by multiplying by $\frac{\sqrt{8} }{\sqrt{8} }$ (which is equal to one so we can use it) which is equal to $\frac{-7\sqrt{8} }{8}$ . Let's finish this by simplifying it. $\sqrt{8} =2\sqrt{2}$ (2x $2^{2}$ ). We can simplify it further by simplifying the 2, making it $\frac{-7\sqrt{2} }{4}$ .