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NeTakaya
3 years ago
10

What are two solutions to x^2+4x+20=0

Mathematics
1 answer:
Zigmanuir [339]3 years ago
4 0
I think the two solutions would be
x =  - 2(1 - 2i) \: and \:  - 2(1 + 2i)

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3/4x - 6y = 18<br>what's the x intercepts and y intercepts?​
mash [69]

Answer:

X: (24,0)

Y(0,-3)

Step-by-step explanation:

X-intercept:

.75x-6y=18

plug in 0 for y

.75x-6(0)=18

divide by .75

x=24

(24,0)

Y-intercept:

plug in 0 for x

.75(0)-6y=18

divide by -6

y=-3

(0,-3)

7 0
3 years ago
(5 pts) A quadratic function f is given by f(x) = ax2 + bx + c where a is not 0. Select all
Vsevolod [243]

Answers:

  • a) True. Plug in x = 0 and it leads to y = c. Therefore, the point (0,c) is on the parabola.
  • b) False. Plug in y = 0 and apply the quadratic formula. One x intercept may sometimes be x = c, but it could easily be other values as well. Or perhaps you may not get any real number solutions at all. See part d) below.
  • c) True. If a < 0, then the leading coefficient is negative. Overall, both endpoints will tend toward negative infinity to produce a parabola that opens downward.
  • d) False. The quadratic may have one x intercept or it may not have any x intercepts at all. It depends on what the discriminant d = b^2 - 4ac is equal to. If d < 0, then we have no x intercepts. If d = 0, then we have exactly 1 x intercept. If d > 0, then we have two different x intercepts.
  • e) True. The vertex's x coordinate is -b/(2a). If b = 0, then the x coordinate of the vertex is 0. The vertical line x = 0 is directly on top of the y axis.
8 0
3 years ago
What is the completely factored form of f(x)=x3+5x2+4x−6?
Gre4nikov [31]
f(x)=x^3+5x^2+4x-6
f(-3)=(-3)^3+5(-3)^2+4(-3)-6=0\implies x+3\text{ is a factor of }f(x)

Synthetic division yields

-3  |  1   5   4   -6
.    |      -3  -6    6
- - - - - - - - - - - - -
.    |  1   2  -2    0

which translates to

\dfrac{x^3+5x^2+4x-6}{x+3}=x^2+2x-2

with remainder 0. Now by the quadratic formula,

x^2+2x-2=0\implies x=\dfrac{-2\pm\sqrt{2^2-4(1)(-2)}}2=-1\pm\sqrt3

and so

f(x)=x^3+5x^2+4x-6=(x+3)(x-(-1+\sqrt3))(x-(-1-\sqrt3))
3 0
3 years ago
Which inequality has no solution?<br> 6(x+2) &gt; x-3<br> 3+4x2(1+ 2x)<br> -2(x+6) X-9 &lt;3(x-3)
Juli2301 [7.4K]

Answer:

6(x + 2) > x - 3

6x + 12 >  x - 3

6x - x >  - 12 - 3

5x >  - 9

3 + 4 \times 2(1 + 2x)

3 + 8(1 + 2x)

3 + 8 + 16x

11 + 16x

- 2(x + 6)x - 9 < 3(x - 3)

- 2( {x}^{2}  + 6x) - 9 < 3x - 9

- 2 {x}^{2}  - 12x - 9 < 3x - 9

5 0
2 years ago
How to rewrite -20/5
alina1380 [7]

Answer:

-4

Step-by-step explanation:

Simplify the equation by dividing both numbers by 5. You will get -4/1. A 1 as a denominator is usually ignored.

5 0
2 years ago
Read 2 more answers
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