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3 years ago

How many zeros does the function ????(x) = 3x^2 + 6x + 2 have? Explain how you know.

Mathematics

1 answer:

maxonik [38]3 years ago

6 0

Answer:

The function has 2 zeros:

$x = -1 - \frac{1}{\sqrt{3}}\\x = -1 + \frac{1}{\sqrt{3}}$

Step-by-step explanation:

When you have a polynomial function, you can identify the grade of the function (the grade is the maximum potency) and this is the number of zeros that the function will have. In this case, you have a second-grade polynomial function that has 2 zeros.

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What is 3h^2+6h-18x as a product of two factors?? Please help,*a lot of points*

Vaselesa [24]

-3 is one of them hope this helps

4 0

2 years ago

If k-9=5(k-3)find k

jok3333 [9.3K]

Answer:

K = 3/2

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

k−9=5(k−3)

k+−9=(5)(k)+(5)(−3)(Distribute)

k+−9=5k+−15

k−9=5k−15

Step 2: Subtract 5k from both sides.

k−9−5k=5k−15−5k

−4k−9=−15

Step 3: Add 9 to both sides.

−4k−9+9=−15+9

−4k=−6

Step 4: Divide both sides by -4.

-4k/-4 = -6/-4

K = 3/2

7 0

3 years ago

Read 2 more answers

A parabola has a focus of F(2, -0.5) and a directrix of y=-1.5 P(x,y) represents any point on the parabola, while D(x, -1.5) rep

prohojiy [21]

The sketch of the parabola is attached below

We have the focus $(a,b) = (2, -0.5)$
The point $P(x,y)$
The directrix, c at $y=-1.5$

The steps to find the equation of the parabola are as follows

Step 1
Find the distance between the focus and the point P using Pythagoras. We have two coordinates; $(2, -0.5)$ and $(x,y)$ .
We need the vertical and horizontal distances to find the hypotenuse (the diagram is shown in the second diagram).
The distance between the focus and point P is given by
$\sqrt{ (x-a)^{2}+ (y-b)^{2} }$

Step 2
Find the distance between the point P to the directrix $c$ . It is a vertical distance between y and c, expressed as $y-c$

Step 3
The equation of parabola is then given as
$\sqrt{ (x-a)^{2}+ (y-b)^{2} }$ = $y-c$
$(x-a)^{2}+ (y-b)^{2}= (y-c)^{2}$ ⇒ substituting a, b and c
$(x-2)^{2}+ (y--0.5)^{2} = (y--1.5)^{2}$
$(x-2)^{2}+ (y+0.5)^{2}= (y+1.5)^{2}$ ⇒Rearranging and making $y$ the subject gives

$y= \frac{ x^{2} }{2} -2x+1$