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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Surface Area of Cube = width^2 * 6 (sides of the cube)
Therefore, (3.5)^2 * 6 =
73.5 [whatever units]^2
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Step-by-step explanation:
You could factor it as 2(x + 3)
9514 1404 393
Answer:
12.0 cm
Step-by-step explanation:
The Pythagorean theorem applies:
(12 cm)² +b² = (17 cm)²
b² = (289 -144) cm² = 145 cm²
b = √145 cm ≈ 12.04 cm
b ≈ 12.0 cm . . . . rounded to 1 decimal place
