Q: Which point is outside the graph of Y multiply 5?

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If x equals a constant number, the graph will be a vertical line. For example, the graph of x = 5 would be a vertical line that goes through the point (5,0). x equals 5 on every point along this lines.

The figures are exactly the same, but every point on the first graph is exactly 13 below the corresponding point on the second one.

1) You write the equation in slope-intercept form, if it isn't in that form already. 2) An easy way to graph it is to start with the y-intercept. For example, if the intercept is +5, you graph the point (0, 5). Then you add an additional point, according to the slope. For example, if the slope is 1/2, you go 2 units to the right, and one up, and graph a point there.

multiply each coordinate by -1For Example:Starting coordinates ---> (5,3)Multiply by -1 ---> (5,3) * -1Final answer ---> (-5,-3)

i think you should have said y=5x which can have different answers for slope intercept graph you start at the origin,move 5 points up and run 1 point

Related questions

5/4 is a just a single point on a numbered line. That point is its graph.

10,10

10,0

If x equals a constant number, the graph will be a vertical line. For example, the graph of x = 5 would be a vertical line that goes through the point (5,0). x equals 5 on every point along this lines.

The figures are exactly the same, but every point on the first graph is exactly 13 below the corresponding point on the second one.

1) You write the equation in slope-intercept form, if it isn't in that form already. 2) An easy way to graph it is to start with the y-intercept. For example, if the intercept is +5, you graph the point (0, 5). Then you add an additional point, according to the slope. For example, if the slope is 1/2, you go 2 units to the right, and one up, and graph a point there.

multiply each coordinate by -1For Example:Starting coordinates ---> (5,3)Multiply by -1 ---> (5,3) * -1Final answer ---> (-5,-3)

The graph of x=5 would simply be a vertical line that passes through the x-intercept of 5. Since x=5, any single point on the graph would have the x coordinate of 5, because no matter what, x=5. This is why It is a vertical line through (5,0).

5% is 5/100. So you can divide by 100 by moving the decimal point 2 places to the left and then multiply by 5, or you can multiply by 5 first and then move the decimal point. Or you can divide by 2 and move the decimal point 1 place to the left. Either way, I get 9.525.

When y=4, x=8/5

Before you start with limits, you should know that they are quite similar to finding the instantaneous rate of change. The limit of any given point (a) on the graph of a function would be the value the graph converges to at that point. The limit, in other words, is the slope of the tangent at a certain point on the graph. For example, take the graph of y = x [Which is the same as f(x) = x] Now, when you graph that function you get a perfectly diagonal line. You can just start at the point (0,0) on the graph and then for each point, go up 1, right 1. Do the same for the left part of the graph, going down 1 and left 1. Now that you got the graph, take ANY value of x. Say you take 5. Now what point is your FUNCTION approaching from EACH side. So its clear that your function is approaching a value of 5 on the y-axis when x=5, from each side i.e. the graph approaches 5 on the y-axis from the left and the right when x =5. Remember that for a limit to exist, the graph should always approach a certain point from BOTH directions, left and right. Consider the graph of y=x2. At x =5, y = 25. Now since the graph approaches the point 25, when x = 5 from both left and right sides, the limit as the graph approaches x=5 is 25!! Remember that it does NOT matter if the graph is defined at the point at which you are finding if the limit exists, what only matters is if the graph is approaching the point from both sides. So to say, you can have a hole at (5,25) and still have the limit as 25. Now there's a specific way of writing limits. Have a look at this image: http://upload.wikimedia.org/math/e/8/7/e879d1b2b7a9e19d16438c24fb8a7990.png Okay, I'll describe what the image states. All its saying is that as x approaches point 'p' on the function f(x), the limit is L. So, to say for the example I just did above, you have have '5' instead of 'p', and 'L' would be replaced by '25'. Now, say the limit at x=2, for the function f(x) is 10, but you actually have a hole at the point (2,10). And you have a DEFINED point at (2,12). IF your graph is still approaching the hole at (2,10) from both sides, then your limit will still exist. Moving on, suppose a point is x = 3 on a certain graph. So, in 'calculus terms', when the graph is approaching 3 from the left side it would be written like 3- while approaching from the right would be 3+.

You could if you want to, but there's not much point to it.