When we do a reflection over a given axis, the distance between the initial point to the axis must be the same as the distance of the reflected point to the axis.

So if we do a reflection over the y-axis, then the value of y must be fixed.

So if we start with the point (x, y), the only other point that is at the same distance from the y-axis is the point (-x, y)

So the rule is, the y value remains equal and the x changes of sign.

4 0

3 years ago

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The simplified expression

Romashka-Z-Leto [24]

Answer:

$5x^2 y^2$

Step-by-step explanation:

We need to use the properties shown below to solve this:

1. $\sqrt[n]{x^a} =x^{\frac{a}{n}}$

2. $\sqrt{x}\sqrt{x} =x$

3. $\sqrt{x} \sqrt{y}=\sqrt{x*y}$

Area of a triangle is given by 1/2 * base * height, so we do that and simplify:

$A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} } y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2$

6 0

3 years ago

Which of the following is a linear function?

Amiraneli [1.4K]

I got y = two over three x

6 0

3 years ago

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