SOLUTION: How would this equation look on a graph?
y= 7/4x + 4
The 7/4 is a fraction, and the x is meant to go with it.
Algebra.Com
Question 456447: How would this equation look on a graph?
y= 7/4x + 4
The 7/4 is a fraction, and the x is meant to go with it.
Found 3 solutions by Alan3354, rfer, josmiceli:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
It's a straight line.
It passes thru (0,4) and it increases 7 units of y for each 4 units of x.
dl the FREE graph software at
http://www.padowan.dk
and look at it.
----------
PS 7x/4 or (7/4)x would be less ambiguous.
Answer by rfer(16322) (Show Source): You can put this solution on YOUR website!
the 7/4 is the slope
4 is the y intercept (0,4)
Start graph at (0,4), then go right 4 and up 7 for second point, and so on.
In the 7/4, the 4 is the x axis and the 7 is the y axis
The y axis is always on the top.
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Here's the graph:
It has a positive slope equal to
and a y-intercept =
RELATED QUESTIONS
y=x2+4x+3 Graph using Quadratic Equation. I have a calculator but I really don't know... (answered by rfer)
How do I solve graphically for x and y? The equation is 3x-y=-4 and 2x-3y=9. How would it (answered by stanbon)
Okay so I have this equation:
h(x) = x^2 - 1
.........---------------... (answered by josgarithmetic)
The problem is... Write an equation for a linear function whose graph has the given... (answered by ankor@dixie-net.com)
#3
what does this look like in a graph after being plotted out?
First, set Y to 0 then... (answered by Alan3354)
Using graph paper, graph each question and plot the given ordered pair. Then construct a... (answered by ankor@dixie-net.com)
Graph the equation using the slope and y-intercept.
y = 9/4x + 7 (the 9/4 is a... (answered by rapaljer)
This review will it ever end. I kinda need a refresher on solution sets and how to graph (answered by stanbon)
I am trying to solve a linear equation with three variables and I am stuck on this... (answered by KMST)