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timama [110]
3 years ago
13

Find the roots. 2x^2 + 4x - 5 =0

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
8 0

Answer:

-1\pm\dfrac{\sqrt{14}}{2}

Step-by-step explanation:

2x^2+4x-5=0

Using the quadratic formula:

x=\dfrac{-4\pm \sqrt{16+40}}{4}=-1\pm\dfrac{\sqrt{14}}{2}

Hope this helps!

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A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and
goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}

Solve for S(t):

\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}

e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}

The left side is the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}

Integrate both sides:

e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt

e^{t/800}S(t)=64e^{t/800}+C

S(t)=64+Ce^{-t/800}

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

0=64+C\impleis C=-64

and so (b) the amount of sugar in the tank at time t is

S(t)=64\left(1-e^{-t/800}\right)

As t\to\infty, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0
3 years ago
14,37,60,83,.... What's the rest of the patern?
Ira Lisetskai [31]
106, 129, 152, 175, 198 count by 23
6 0
3 years ago
Read 2 more answers
A system for tracking ships indicated that a ship lies on a hyperbolic path described by 5x2 - y2 = 20. the process is repeated
zysi [14]
Answer:
The ship is located at (3,5)

Explanation:
In the first test, the equation of the position was:
5x² - y² = 20 ...........> equation I
In the second test, the equation of the position was:
y² - 2x² = 7 ..............> equation II
This equation can be rewritten as:
y² = 2x² + 7 ............> equation III

Since the ship did not move in the duration between the two tests, therefore, the position of the ship is the same in the two tests which means that:
equation I = equation II

To get the position of the ship, we will simply need to solve equation I and equation II simultaneously and get their solution.

Substitute with equation III in equation I to solve for x as follows:
5x²-y² = 20
5x² - (2x²+7) = 20
5x² - 2y² - 7 = 20
3x² = 27
x² = 9
x = <span>± </span>√9

We are given that the ship lies in the first quadrant. This means that both its x and y coordinates are positive. This means that:
x = √9 = 3

Substitute with x in equation III to get y as follows:
y² = 2x² + 7
y² = 2(3)² + 7
y = 18 + 7
y = 25
y = +√25
y = 5

Based on the above, the position of the ship is (3,5).

Hope this helps :)
8 0
3 years ago
Which pair of fractions is equivalent to this pair?
I am Lyosha [343]
The answer is B 10/18 and 3/18
8 0
3 years ago
Write the following as an inequality 6 is greater than or equal to x and -2 is less than or equal to x
kipiarov [429]

Answer:

  -2 ≤ x ≤ 6

Step-by-step explanation:

The first is ...

  6 ≥ x

and it can also be written as ...

  x ≤ 6

The second is

  -2 ≤ x

The two inequalities can be combined into one compound inequality:

  -2 ≤ x ≤ 6

_____

<em>Comment on this answer</em>

I prefer the answer in this form because it puts the parts of the expression in the same order that they appear on a number line. It can also be written as ...

  6 ≥ x ≥ -2

7 0
3 years ago
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