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:
about a half teaspoon
Step-by-step explanation:
Answer:
y = 5/6x - 5.167
Or
y = 5/6x - 31/6
Step-by-step explanation:
The equation of a line
y = mx + c
Step 1: find the slope
m = y2 - y1 / x2 - x1
Given points
( 0 , -7) ( 5 , -1)
x1 = 0
y1 = -7
x2 = 5
y2 = -1
Insert the values into the equation
m = -1 - (-7) / 5 - 0
m = -1 + 7 / 5
m = 6/5
y = 6/5x + c
Step 2: sub any of the two points given into the equation
y = 5/6x + c
( 5 , -1)
x = 5
y = -1
-1 = 5/6(5) + c
-1 = 5*5/6 + c
-1 = 25/6 + c.
c = -1 - 25/6
LCM = 6
c = -6 - 25 / 6
c = -31/6
c = -5.167
Step 3: sub c into the equation
y = 5/6x + c
y = 5/6x - 31/6
Or
y = 5/6x - 5.167
The equation of the line is
y = 5/6x - 5.167
Answer:
Each pair costs $8.86
Step-by-step explanation:
Divide 35.45 by 4
You get 8.8625
Round that to the nearest hundredth.
You get $8.86
Hope this helped!
