Write an equation of a line that is parallel to the line whose equation is 3y=x+6 and that passes through the point (-3,4).

1 answer:

3 0

The standard form of this equation is

$y=\frac{1}{3}x+2$

The slope of two parallel lines i the same, so the slope of the line we're looking for is also $\frac{1}{3}$

Since the line passes through $(-3,4)$ , the equation is:

$(y-4)=\frac{1}{3}(x+3)$

$\boxed{y=\frac{1}{3}x+5}$

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The figure below is a square. Find the length of side x in simplest radical form with a rational denominator.

Mashcka [7]

Answer:

$\frac{\sqrt{10}}{2}$

Step-by-step explanation:

The diagonal forms two 45-45-90 triangles, with the diagonal being the hypotenuse of both. The Pythagorean Theorem states that $a^2+b^2=c^2$ , where $c$ is the hypotenuse of the triangle, and $a$ and $b$ are the two legs of the triangle.

From the Isosceles Base Theorem, the two legs of a 45-45-90 triangle are always equal. Since we're given a diagonal of $\sqrt{5}$ , we have:

$x^2+x^2=\sqrt{5}^2,\\2x^2=5,\\x^2=\frac{5}{2},\\x=\sqrt{\frac{5}{2}}=\frac{\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{5}}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\boxed{\frac{\sqrt{10}}{2}}$