Pls I need help ASAP <br><br> What is 2.5

If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?

Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

$(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2$

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

5 0

3 years ago

A projectile is fired with muzzle speed 220 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

Allushta [10]

Answer:

4968.6 m from where it was fired
221.33 m/s

Step-by-step explanation:

For the purpose of this problem, we assume ballistic motion over a stationary flat Earth under the influence of gravity, with no air resistance.

We can divide the motion into two components, one vertical and one horizontal. For muzzle speed s and launch angle θ, the horizontal speed is presumed constant at s·cos(θ). The initial vertical speed is then s·sin(θ) and the (x, y) coordinates as a function of time are ...

(x, y) = (s·cos(θ)·t, -4.9t² +s·sin(θ)·t + h₀) . . . . . where h₀ is the initial height

To find the range, we can solve the equation y=0 for t, and use this value of t to find x.

Using the quadratic formula, we find t at the time of landing to be ...

t = (-s·sin(θ) - √((s·sin(θ))²-4(-4.9)(h₀)))/(2(-4.9))

t = (s/9.8)(sin(θ) +√(sin(θ)² +19.6h₀/s²))

For s = 220, θ = 45°, and h₀ = 30, the time of flight is ...

t ≈ 31.939 seconds

Then the horizontal travel is

x = 220·cos(45°)·31.939 ≈ 4968.6 . . . . meters

__

As it happens, the value under the radical in the above expression for time, when multiplied by s, is the vertical speed at landing. The horizontal speed remains s·cos(θ), so the resultant speed is the Pythagorean sum of these:

landing speed = s·√(cos(θ)² +sin(θ)² +19.6h₀/s²) ≈ s√(1 +0.012149)

≈ 221.33 m/s

_____

Note that the landing speed represents the speed the projectile has as a consequence of the potential energy of its initial height being converted to kinetic energy that adds to the kinetic energy due to its initial muzzle velocity.

6 0

2 years ago

Fiona y sus amigas fueron a una tienda de batidos. Pidieron tres batidos de fresa y dos batidos de proteínas por un total de $ 2

jeka94

Answer:

Un batido de proteínas cuesta $ 5,25.

Step-by-step explanation:

El sistema de ecuaciones descrito al final del enunciado coincide con lo descrito al comienzo del mismo, de modo que podemos concentraron casi exclusivamente en la resolución del sistema en cuestión, la cual se describe a continuación:

Este es el sistema a resolver:

$x + 4\cdot y = 25,50$ (1)