Which linear inequality is represented by the graph?
1 answer:
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<h2>Explanation:</h2>Since you don't provide any graph I can't give you an exact answer. However, I'll give you the fourth graph for this problem. So we have:
<h3>First:</h3>We have the inequality:
The symbol < tells us that the line is dotted and the shaded region is below the line. Also, the slope is 3 and the y-intercept is 2. So as x increases one unit y increases 3, so if (0, 2) which is both a point and the y-intercept of the line, then (1, 5) will also be a point on the line. By using graphing tool we get the first graph below.
<h3>Second:</h3>We have the inequality:
This case is similar to the previous one. The only difference is that the shaded region is above the line. The graph is the second one below.
<h3>Third:</h3>We have the inequality:
Again the line is shaded under the graph. Also, the slope is 1/3 and the y-intercept is 2. So as x increases one unit y increases 1/3, or in other words, as x increase 3 units y increases one unit, so if (0, 2) which is both a point and the y-intercept of the line, then (3, 3) will also be a point on the line. By using graphing tool we get the third graph below.
<h3>Fourth:</h3>We have the inequality:
This case is similar to the previous one. The only difference is that the shaded region is above the line. The graph is the fourth one below.
<h2>Learn more:</h2>Inequalities: brainly.com/question/12890742
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The smallest such number is 1055.
We want to find such that
The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.
Taking everything together, we end up with the system
Now the moduli are coprime and we can apply the CRT.
We start with
Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.
Taken modulo 2, we end up with
which means the first term is fine and doesn't require adjustment.
Taken modulo 3, we have
We want a remainder of 2, so we just need to multiply the second term by 2.
Taken modulo 5, we have
We want a remainder of 0, so we can just multiply this term by 0.
Taken modulo 7, we have
We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since , the inverse of 2 is 4.
So, we have to adjust to
and from the CRT we find
so that the general solution for all integers
.
We want a 4 digit solution, so we want
which gives .