In order to determine whether the equations are parallel, perpendicular, or neither, let's simply each equation into a slope-intercept form or basically, solve for y.

Let's start with the first equation.

$\frac{6x-5y}{2}=x+1$

Cross multiply both sides of the equation.

$6x-5y=2(x+1)$ $6x-5y=2x+2$

Subtract 6x on both sides of the equation.

$6x-5y-6x=2x+2-6x$ $-5y=-4x+2$

Divide both sides of the equation by -5.

$-\frac{5y}{-5}=\frac{-4x}{-5}+\frac{2}{-5}$ $y=\frac{4}{5}x-\frac{2}{5}$

Therefore, the slope of the first equation is 4/5.

Let's now simplify the second equation.

$-4y-x=4x+5$

Add x on both sides of the equation.

$-4y-x+x=4x+5+x$ $-4y=5x+5$

Divide both sides of the equation by -4.

$\frac{-4y}{-4}=\frac{5x}{-4}+\frac{5}{-4}$ $y=-\frac{5}{4}x-\frac{5}{4}$

Therefore, the slope of the second equation is -5/4.

Since the slope of each equation is the negative reciprocal of each other, then the graph of the two equations is perpendicular to each other.

5 0

1 year ago

What is the common denominator as a fraction of 1/5

andrezito [222]

The answer would be 30 if u use the gcf

4 0

3 years ago

Read 2 more answers

a trapezoide has an area of 150 square meters. If the bases are 14 meters and 16 meters, what is the height of the trapezoid

SIZIF [17.4K]

Answer:

h = 10 m

Step-by-step explanation:

The formula for area of a trapezoid is

A = (1/2)(B + b)h where B is the bottom base, b is the top base, and h is the height.

We are given B = 16, b = 14, and A = 15, plug those values in and simplify

150 = (1/2)(16 + 14)(h)

150 = (1/2)(30)(h)

150 = (15)h (half of 30 is 15)

10 m = h (divide both sides by 15 to isolate h)

6 0

3 years ago