Answer: The value of k for which one root of the quadratic equation kx2 - 14x + 8 = 0 is six times the other is k = 3.
Let's look into the solution step by step.
Explanation:
Given: A quadratic equation, kx2 - 14x + 8 = 0
Let the two zeros of the equation be α and β.
According to the given question, if one of the roots is α the other root will be 6α.
Thus, β = 6α
Hence, the two zeros are α and 6α.
We know that for a given quadratic equation ax2 + bx + c = 0
The sum of the zeros is expressed as,
α + β = - b / a
The product of the zeros is expressed as,
αβ = c / a
For the given quadratic equation kx2 - 14x + 8 = 0,
a = k, b = -14, c = 8
The sum of the zeros is:
α + 6α = 14 / k [Since the two zeros are α and 6α]
⇒ 7α = 14 / k
⇒ α = 2 / k --------------- (1)
The product of the zeros is:
⇒ α × 6α = 8 / k [Since the two zeros are α and 6α]
⇒ 6α 2 = 8 / k
⇒ 6 (2 / k)2 = 8 / k [From (1)]
⇒ 6 × (4 / k) = 8
⇒ k = 24 / 8
⇒ k = 3
Answer:
A. Plane B because it was 9.33 miles away
B. 48 units
Step-by-step explanation:
A. Since the airplanes fly at an angle to the runway, their direction forms a triangle with the runway with their height above the ground as the opposite of the angle and their distance from the airport as the hypotenuse.
So for airplane A with 44° angle of departure,
sin44° = y/h where y = height above the ground and h = distance from airport
So h = y/sin44° = 6/sin44° = 8.64 miles
So for airplane B with 40° angle of departure,
sin40° = y/H where y = height above the ground and H = distance from airport
So H = y/sin40° = 6/sin40° = 9.33 miles
Since airplane B is at 9.33 miles away from the airport whereas airplane A is 8.64 miles from the airport, airplane B is farther away.
B. We know that scale factor = new size/original size
Our scale factor = 4 and original size = 12 units. So,
new size = scale factor original size = 4 × 12 = 48 units.
Sqrt(-2), can be written as i * sqrt(2). Thus, our answer is x - i sqrt(2).
66cm is equal to 6.6e+8 nanometers.
Answer:
what? what is this supposed to be?
