Six times a number increased by 4 is no less than 50. Which inequality represents this statement?
2 answers:
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The solution is the point of intersection between the two equations.
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution
The four?
Of course there are an infinite number of rationals between any two different real numbers, as well as an even bigger infinite number of irrationals.
The average will be between the numbers:
That's a lot of work to get a number in between. We can just see
is between the two numbers.
The mediant, or freshman addition, will always be in between:
2/5=.4 and 5/6 is about .83, so
is in between as is .7, .41, .530940394 and as many rationals as we care to generate.
To get irrationals we could just add a teeny irrational to the ones we just generated, like
We could just change that denominator a bit and get as many as we like.
But let's get some square roots. The geometric mean will be between
That's the tangent from one of trig's biggest cliches, but I digress. It's in between.
While we're on trig cliches,
is in between as well.
Keeping with the trig theme, also in between are
and
the angles associated with the some of the above trig function values.
We could obviously go on as long as we cared to.
Of course there are an infinite number of rationals between any two different real numbers, as well as an even bigger infinite number of irrationals.
The average will be between the numbers:
That's a lot of work to get a number in between. We can just see
is between the two numbers.
The mediant, or freshman addition, will always be in between:
2/5=.4 and 5/6 is about .83, so
is in between as is .7, .41, .530940394 and as many rationals as we care to generate.
To get irrationals we could just add a teeny irrational to the ones we just generated, like
We could just change that denominator a bit and get as many as we like.
But let's get some square roots. The geometric mean will be between
That's the tangent from one of trig's biggest cliches, but I digress. It's in between.
While we're on trig cliches,
is in between as well.
Keeping with the trig theme, also in between are
and
the angles associated with the some of the above trig function values.
We could obviously go on as long as we cared to.