The equation in slope-intercept form represents a line that is parallel to y = 12x − 2 and passes through the point (−8,1) is y = 12x + 97
<em><u>Solution:</u></em>
Given that we have to write the equation in slope intercept form for line that is parallel to y = 12x − 2 and passes through the point (−8, 1)
<em><u>The slope intercept form is given as:</u></em>
y = mx + c ---- eqn 1
Where "m" is the slope of line and "c" is the y - intercept
Given equation of line is y = 12x - 2
On comparing the above equation with eqn 1,
m = 12
Thus slope of line is 12
We know that slopes of parallel lines are equal
Therefore, slope of line parallel to given line is also 12
Given point is (-8, 1)
We have to find the equation of line passing through (-8, 1) with slope m = 12
Substitute m = 12 and (x, y) = (-8, 1) in eqn 1
1 = 12(-8) + c
1 = -96 + c
c = 97
Substitute c = 97 and m = 12 in eqn 1
y = 12x + 97
Thus the required equation of line is found
Answer: 150 pounds
Step-by-step explanation:
Let her initial weight be represented by x.
Since Shep had lost 16% of his weight, which was 24 pounds. Shep weight before he was put on a diet would be:
16% × x = 24
0.16 × x = 24
0.16x = 24
x = 24/0.16
x = 150 pounds
Answer:
10 inches
Step-by-step explanation:
18*2=36
56-36=20
20/2= 10
The missing side length is 10 inches.
Hope this helps! Have a nice day :)
Answer: y=2x+1 is parallel & y=-2x+5 is perpendicular
Answer:
If a line is perpendicular to another line, that means that the slope is completely opposite that of the original line. The first thing that we do to the slope is we negate the number which means that if we have a slope of 
our slope because 
in this step. In our case our slope is so in this step it becomes .
Moving onto the second part which is to get the reciprocal of the number which means that if we have 
then we would switch it to 
. In our case our number is so we would make that into a fraction like this 
.
In conclusion, our final slope of the perpendicular line is .
<u><em>Hope this helps! Let me know if you have any questions</em></u>
