The correct answer is b, x > 1.
To find the domain of a square root function, we need to ensure that the value inside the square root is non-negative. In this case, we have y = square root(3 + 3), so 3 + 3 = 6. Since 6 is greater than 0, the domain of this function is all real numbers greater than 1.
What is domain of the function? y = sqaure root 3+ 3
a, x < -1
b, x > 1
c, x > -1
d, x < 1
3 answers
Simplify the rational expression by rationalizing the denominator.
sqaure root 15 sqaure root 6x
sqaure root 15 sqaure root 6x
To simplify the rational expression \(\frac{\sqrt{15}\sqrt{6x}}{1}\), we want to rationalize the denominator.
First, multiply the numerator and denominator by \(\sqrt{6x}\) to rationalize the denominator:
\(\frac{\sqrt{15}\sqrt{6x}\cdot \sqrt{6x}}{1\cdot \sqrt{6x}} = \frac{\sqrt{15}\cdot 6x}{\sqrt{6x}} = \frac{6x\sqrt{15}}{\sqrt{6x}}\)
To simplify further, we can rewrite \(\sqrt{6x}\) as \(\sqrt{6}\sqrt{x}\):
\(\frac{6x\sqrt{15}}{\sqrt{6}\sqrt{x}} = \frac{6\sqrt{15}x}{\sqrt{6}\sqrt{x}} = \frac{6\sqrt{15}x}{\sqrt{6x}}\)
Therefore, the simplified expression after rationalizing the denominator is \(\frac{6\sqrt{15}x}{\sqrt{6x}}\).
First, multiply the numerator and denominator by \(\sqrt{6x}\) to rationalize the denominator:
\(\frac{\sqrt{15}\sqrt{6x}\cdot \sqrt{6x}}{1\cdot \sqrt{6x}} = \frac{\sqrt{15}\cdot 6x}{\sqrt{6x}} = \frac{6x\sqrt{15}}{\sqrt{6x}}\)
To simplify further, we can rewrite \(\sqrt{6x}\) as \(\sqrt{6}\sqrt{x}\):
\(\frac{6x\sqrt{15}}{\sqrt{6}\sqrt{x}} = \frac{6\sqrt{15}x}{\sqrt{6}\sqrt{x}} = \frac{6\sqrt{15}x}{\sqrt{6x}}\)
Therefore, the simplified expression after rationalizing the denominator is \(\frac{6\sqrt{15}x}{\sqrt{6x}}\).
1 answer