Find the standard form of equation of a line perpendicular to 4x plus 5y equals 40 that includes the point -10, -3

The equation of line in standard form is 5x - 4y = -38

Solution:

Given that we have to find the equation of line perpendicular to 4x + 5y = 40 that includes the point (-10, -3)

The equation of line in slope intercept form is given as:

y = mx + c ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Given equation of line is:

4x + 5y = 40

Rearranging to slope intercept form, we get

5y = -4x + 40

On comparing the above equation with eqn 1,

We know that product of slope of line and slope of line perpendicular to given line is equal to -1

Therefore,

Now find the equation of line with slope 5/4 and passing through (-10, -3)

Substitute and (x, y) = (-10, -3) in eqn 1

Substitute and in eqn 1

The standard form of an equation is Ax + By = C

Therefore,

Thus the equation of line in standard form is found

38.5

step-by-step explanation:

lets say, first number = r

second number = 3r+7

total = 49

(r)+(3r+7)= 49

4r+7 = 49

4r = 42

r = 42/4

r= 10.5

3r+7 = 3(10.5)+7 = 38.5

hope this

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.