Dy/dx=1/2, 1/2, 1/2 etc
So this is a linear equation of the form y=mx+b where m=dy/dx=1/2 so
y=x/2 +b, now we can use any point to solve for the y-intercept, "b", I'll use (7,0)
0=7/2 +b
b=-7/2 so
y=x/2-7/2
y=(x-7)/2
Answer:
third one
Step-by-step explanation:
Answer:
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Find a point-slope form for the line that satisfies the stated conditions. Slope , passing through (-5,4)
I really need this question answered
By:
I don't see a value for the slope. We need that to set the equation, otherwise I can write an unlimited number of equations that pass through (-5,4).
I'll assume a slope so that you can see how the procedure would work. I like 6, so we'll assume a slope of 6.
The equation for a straight line has the form y = mx + b, where m is the slope and y is the y-intercept, the value of y when x = 0. We want a line that has slope 6, so:
y = 6x + b
We need to find b, so substitute the point (-5,4) that we know is on the line:
4 = 6*(-5) + b and solve for b
4 = -30 + b
b = 34
The line is y = 6x + 34
Answer:
given us,
(x1y1) = (4,6)
(x2y2)= (1,10)
here
slope= (y2-y1)/ (x2-x1)
= (10-6)/ (1-4)
= 4/ (-3)
Step-by-step explanation:
4/ (-3)
Answer:x= -8 or 2
Step-by-step explanation:
