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aivan3 [116]
3 years ago
13

X^2 + 5x = quadratic equation =​

Mathematics
1 answer:
timama [110]3 years ago
4 0

Hi there!

The question gives us the quadratic equation , and it tells us to solve it using the quadratic formula, which goes as . However, we must first find the values of a, b, and c. The official quadratic equation goes as , which matches the format of the given quadratic equation. Hence, the value of a would be 1, the value of b would be 5, and the value of c would be 3. Now, just plug it back into the quadratic equation and simplify to get the zeros of the equation. 

x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}  

x = \frac{-(5) \pm \sqrt{(5)^2 - 4(1)(3)} }{2(1)}  

x = \frac{-5 \pm \sqrt{25 - 12} }{2}  

x = \frac{-5 \pm \sqrt{13} }{2}  

x = \frac{-5 \pm 3.61 }{2}  

x = \frac{-5 + 3.61 }{2}, x = \frac{-5 - 3.61 }{2}

x=-0.695 \ \textgreater \ \ \textgreater \  -0.7, x= -4.305 \ \textgreater \ \ \textgreater \ x=-4.31

Therefore, the solutions to the quadratic equation  are x = -0.7 and x = -4.31. Hope this helped and have a phenomenal day!

Your answer is 4.31

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kykrilka [37]

Answer:

A culture started with 5000 bacteria. after 4 hours, it grew to 6000.

=> Increasing rate = difference/time = (6000 - 5000)/4 = 250 bacteria/hour

=> The number of bacteria after 17 hours:

N = 5000 + 250 x 17 = 9250 bacteria

Hope this helps!

:)

4 0
3 years ago
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An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object
Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2  + \frac{1}{2} at^2\\3) v = v_0 + at

What do we already know?

  • v_0 = 29.4 ms^-1
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  • a = -9.81 ms^-2 this is the gravitational acceleration g
  • s_0 = 0 m, since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, v = 0 ms^-2

v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2

this is the equation of the whole trajectory that object makes.

and if you solve this by making s = 0, you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0
3 years ago
Let x=number of cds and y=number of dvds
Elis [28]
X + y = 40.....multiply by -4
4x + 6y = 180
-------------------
-4x - 4y = -160 (result of multiplying by -4)
4x + 6y = 180
------------------add
2y = 20
y = 20/2
y = 10 <===

x = 30

6 0
3 years ago
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Answer:

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Step-by-step explanation:

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Answer:

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