Please hurry I need this soon what is m
1 answer:
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Answer:
(y+7)=5/6(x+6)
Step-by-step explanation:
For this problem we need to use Point-Slope Form to write an equation.
We already have the point, (-6,-7), so all we need now is the slope, which we know is perpendicular to the line 6x+5y=30.
First, let's find the slope of 6x+5y=30 by converting it into Slope-Intercept Form, which is y=mx+b. m is the slope.
6x+5y=30
5y=-6x+30
y=-6/5x+6
So the slope of this line is -6/5, but we need the slope of the line is perpendicular to this line.
When a line is perpendicular to another, their slopes are negative reciprocals to one another, or opposite reciprocals. For example, a line that is perpendicular to another line that has a slope of 3/4 will have a slope of -4/3. Flip the numerator and denominator to find the reciprocal and add a negative.
x
So the negative reciprocal of -6/5 is 5/6, so that is our slope.
Now we can assemble our equation, here is the Point-Slope Form:
again, m is our slope.
and
represent the coordinates of the point that this line passes through, (-6, -7).
Let's plug in the y, x, and m values:
(y-(-7))=5/6(x-(-6))
Simplify:
(y+7)=5/6(x+6)
That is our final answer.
Hello!
Here we can use the quad formula to solve for the x values or roots.
= -2 and 3/2
The negative root for this equation is x=-2
We find the roots of an equation by solving for x by different ways, most common ones being the quadratic formula of x= [-b +/-sqrt(b^2-4ac)]/2a or factoring and setting equal to zero and solving for x value(s) that way. Also the roots are where the graph of the equation crosses the x-axis.
Hope this helps. Any questions please just ask. Thank you kindly
Here we can use the quad formula to solve for the x values or roots.
The negative root for this equation is x=-2
We find the roots of an equation by solving for x by different ways, most common ones being the quadratic formula of x= [-b +/-sqrt(b^2-4ac)]/2a or factoring and setting equal to zero and solving for x value(s) that way. Also the roots are where the graph of the equation crosses the x-axis.
Hope this helps. Any questions please just ask. Thank you kindly