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:
value of x = 5.8 mm
Step-by-step explanation:
We have given,
Two right triangles EDH and EDG.
In right triangle EDH, EH = 56mm , DH = 35 mm
Using Pythagoras theorem we can find ED.
i.e EH² = ED²+DH²
56²=ED²+35²
ED²=56²-35²
ED = √(56²-35²) = 7√39 = 43.71 mm
Now, Consider right triangle EDG
Here, EG=44.8mm , GD = x+4 and ED = 7√39
Again using Pythagoras theorem,
EG² = ED² + DG²
44.8²= (7√39)²+ (x+4)²
(x+4)² = 44.8² - (7√39)²
x+4 = √(44.8² - (7√39)²)
x+4 = 9.8
or x = 9.8 - 4 = 5.8 mm
Hence we got the value of x = 5.8 mm
Answer:
x = 15
Step-by-step explanation:
The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.
∠ KLM is an external angle of the triangle, thus
∠ KLM = ∠ JKL + ∠ KJL , substitute values
5x - 7 = x + 20 + 2x + 3
5x - 7 = 3x + 23 ( subtract 3x from both sides )
2x - 7 = 23 ( add 7 to both sides )
2x = 30 ( divide both sides by 2 )
x = 15
5(2y-4) - 3y = 1. 10y-20-3y=1. 7y-20=1. 7y=21. y =3. x = 2(3)-4. x = 2. x*y =6.
