You end up with a false statement.
4 has never equaled 3.
Hope it helps!
Good lcuk with your studies! :)
Hope it helps! :)
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The solution of the system can be x - 3y = 4 only if both the equations can be simplified to x - 3y = 4.
This will mean that both the equations will result in the same line which is x - 3y = 4 and thus have infinitely many solutions.
Second equation is:
Qx - 6y = 8
Taking 2 common we get:
(Q/2)x - 3y = 4
Comparing this equation to x- 3y = 4, we can say that
Q/2 = 1
So,
Q = 2
Therefore, the second equation will be:
2x - 6y = 8
(t-2)(t+1)(t+1)
t^2+t-2t-2(t+1)
t^2-t-2(t+1)
t^3-t^2-2t+t^2-t-2
t^3+-3t-2
<span>h<span>(t)</span>=<span>t<span>34</span></span>−3<span>t<span>14</span></span></span>
Note that the domain of h is <span>[0,∞]</span>.
By differentiating,
<span>h'<span>(t)</span>=<span>34</span><span>t<span>−<span>14</span></span></span>−<span>34</span><span>t<span>−<span>34</span></span></span></span>
by factoring out <span>34</span>,
<span>=<span>34</span><span>(<span>1<span>t<span>14</span></span></span>−<span>1<span>t<span>34</span></span></span>)</span></span>
by finding the common denominator,
<span>=<span>34</span><span><span><span>t<span>12</span></span>−1</span><span>t<span>34</span></span></span>=0</span>
<span>⇒<span>t<span>12</span></span>=1⇒t=1</span>
Since <span>h'<span>(0)</span></span> is undefined, <span>t=0</span> is also a critical number.
Hence, the critical numbers are <span>t=0,1</span>.
I hope that this was helpful.
Answer:
Yes, 2 is a rational number.
