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:
3.75 quarts, rounded to 3.8 quarts
Step-by-step explanation:
3 quarts of solution is 10% antifreeze, so 0.3 quarts are already antifreeze, and 2.7 quarts are not.
a(the antifreeze we already have)+x(what we're going to add)= 1.5*2.7
Let me explain. If we have 60% antifreeze, 40% is not. 60/40=1.5
Substitute a for 0.3
0.3+x=4.05 Subtract 0.3 from both sides
x=3.75
Add all of the sides together which then equal 76. So it would simplify to 14x+20=76. Then 14x=56. And 56/14=4. X=4
Answer:145,003 59
Step-by-step explanation:
he population 2016, this is more precise, will be 145116 people. If you just rounded the value of X to population growth rate to 1.176, then the population Uh in 2016 is 145,003 59.
Answer:
Step-by-step explanation:
We require 2 equations with the repeating 3 placed after the decimal point.
let x = 0.5333.. ( multiply both sides by 10 and 100 )
10x = 5.333... → (1)
100x = 53.333... → (2)
Subtract (1) from (2) tus eliminating the repeating 3
90x = 48 ( divide both sides by 90 )
x = 
=
