2. Which trinomial below is equivalent to the square of the binomial x-5?
1 answer:
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6x^2 - 2x + 1 is a quadratic formula from the form ax^2 + bx + c. This form of equation represents a parabola.
Finding 6x^2 - 2x + 1 = 0, means that you need to find the zeroes of the equation.
Δ = b^2 - 4ac
If Δ>0, the equation admits 2 zeroes and 6x^2 - 2x + 1 = 0 exists for 2 values of x.
If Δ<0, the equation doesn't admit any zero, and 6x^2 - 2x + 1 = 0 doesn't exist since the parabola doesn't intersect with the axe X'X
If Δ=0, the equation admits 1 zero, which means that the peak of the parabola is touching the axe X'X.
In 6x^2 - 2x + 1, a=6, b=-2, and c =1.
Δ= b^2 - 4ac
Δ=(-2)^2 - 4(6)(1)
Δ= 4 - 24
Δ= -20
Δ<0 so the parabola doesn't intersect with the Axe X'X, which means there's no solution for 6x^2 - 2x + 1 = 0.
I've added a picture of the parabola represented by this equation under the answer.
Hope this Helps! :)
Finding 6x^2 - 2x + 1 = 0, means that you need to find the zeroes of the equation.
Δ = b^2 - 4ac
If Δ>0, the equation admits 2 zeroes and 6x^2 - 2x + 1 = 0 exists for 2 values of x.
If Δ<0, the equation doesn't admit any zero, and 6x^2 - 2x + 1 = 0 doesn't exist since the parabola doesn't intersect with the axe X'X
If Δ=0, the equation admits 1 zero, which means that the peak of the parabola is touching the axe X'X.
In 6x^2 - 2x + 1, a=6, b=-2, and c =1.
Δ= b^2 - 4ac
Δ=(-2)^2 - 4(6)(1)
Δ= 4 - 24
Δ= -20
Δ<0 so the parabola doesn't intersect with the Axe X'X, which means there's no solution for 6x^2 - 2x + 1 = 0.
I've added a picture of the parabola represented by this equation under the answer.
Hope this Helps! :)
Third leg.
The crew flies at a speed of 560 mi/h in direction N-20°-E.
The wind has a speed of 35 mi/h and a direction S-10°-E.
We then can draw this as:
We have to add the two vectors to find the actual speed and direction.
We will start by adding the x-coordinate (W-E axis):
and the y-coordinate (S-N axis) is:
Then, the actual speed vector is v3=(197.61, 491.76).
The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].
Then, we have to calculate the displacement in 20 minutes using the actual speed vector.
We can calculate the movement in each of the axis. For the x-axis:
NOTE: 20 minutes represents 1/3 of an hour.
We can do the same with the y-coordinate:
The final position is R3 = (282.53, 331.59).
To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:
The distance can be calculated as if it was a right triangle:
The angle, from E to N, can be calculated as:
If we want to express it from N to E, we substract the angle from 90°:
Answer: the final location can be represented with the vector (282.53, 331.59).
1) The distance from the origin is 435.63 miles and
2) the direction is N-40°-E.
