The given equation has no solution when K is any real number and k>12
We have given that
3x^2−4x+k=0
△=b^2−4ac=k^2−4(3)(12)=k^2−144.
<h3>What is the condition for a solution?</h3>
If Δ=0, it has 1 real solution,
Δ<0 it has no real solution,
Δ>0 it has 2 real solutions.
We get,
Δ=k^2−144 here Δ is not zero.
It is either >0 or <0
Δ<0 it has no real solution,
Therefore the given equation has no solution when K is any real number.
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Answer:
It will take him 1 hour and 30 minutes. (90 minutes)
Step-by-step explanation:
If he is running 1 mile in 10 minutes and he needs to run 9 miles the just multiply 9 by 10 to get 90 then simplify if you need to to get 1 hour and 30 minutes.
Hope this helped and have a great rest of your day.
If you would like to know how many kilometers did Kelly jog that week, you can calculate this using the following steps:
Morgan: 51.2 kilometers
Kelly: Morgan - 6 kilometers = 51.2 - 6 = 45.2 kilometers
The correct result would be 45.2 kilometers.
Sub (x+1) for x
f(x+1)=2(x+1)^2+3
f(x+1)=2(x^2+2x+1)+3
f(x+1)=2x^2+4x+2+3
f(x+1)=2x^2+4x+5
D is answer
Let e=erasers and p=pencils. So we know the total amount of erasers plus pencils sold is equal to 220. Therefore,
We also know the cost of erasers and pencils totaled to $695 in earnings and so:
and so those are your two equations.
