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

Find the degree of the polynomial<br><br> 7a3b2 – 2a4 + 4b – 15<br><br> Degree

vesna_86 [32]

When ur polynomial has more then one variable....and this one does...the degree is the highest term degree

7a^3b^2 ....this term has a degree of 5
2a^4...this term has a degree of 4
4b....this term has a degree of 1
15....this term has a degree of 0

so the highest degree term has a degree of 5....so that is the degree of this polynomial.

now if u just had 1 variable, the degree would be the highest exponent

3 0

3 years ago