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:
Plot these points:(-2,-7) (-1,-5) (0,-3) (1,-1) (2,1) (3,3) (4,5)
Step-by-step explanation:
You need to substitute your x values into the x of the equation y=2(3)-3 to find the y value...
Answer:
.
Step-by-step explanation:
The linear approximation is given by the equation

Linear approximation is a good way to approximate values of
as long as you stay close to the point
, but the farther you get from
, the worse your approximation.
We know that,

Next, we need to plug in the known values and calculate the value of
:

Then
.
A plane cuts horizontally across a rectangular pyramid will always be a rectangle. (Answer B)
Answer:
B, C, D
Step-by-step explanation:
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