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Bess [88]
3 years ago
6

For the equation y = 2x2 − 5x + 18, choose the correct application of the quadratic formula.

Mathematics
2 answers:
Mazyrski [523]3 years ago
6 0
X equals five plus or minus the square root of negative five squared minus four times two times eighteen, all divided by two times two
gladu [14]3 years ago
3 0

Answer:

Option 2nd is correct

x equals five plus or minus the square root of negative five squared minus four times two times eighteen, all divided by two times two

Step-by-step explanation:

Given the equation:

y = 2x^2-5x+18

Use the quadratic formula:

A quadratic equation y = ax^2+bx+c.............[1], then the solution is given by:

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

On comparing given equation with [1]

a = 2 , b = -5 and c = 18

then;

x = \frac{-(-5) \pm \sqrt{(-5)^2-4(2)(18)}}{2(2)}

⇒x = \frac{5 \pm \sqrt{(-5)^2-4(2)(18)}}{2(2)}

Therefore, the correct application of the quadratic formula is,

x equals five plus or minus the square root of negative five squared minus four times two times eighteen, all divided by two times two

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Answer:

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4 0
3 years ago
Please help me with these questions!! Legit answers only please!!!
zimovet [89]

Answer:

Values of y are:

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6 0
3 years ago
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
The value of y varies inversely as the cube of x and y=1 when x=6. Find the value of y when x =2.
ch4aika [34]

Answer:

y=27 when x=2

Step-by-step explanation:

y varies inversely as the cube of x

y=k(1/X^3)

y=1 when x=6

y=k(1/X^3)

1=k(1/6^3)

1=k(1/216)

k=1÷1/216

K=1×216/1

k=216

find y when x=2

y=k(1/X^3)

y=216(1/2^3)

y=216(1/8)

y=216/8

y=27

3 0
3 years ago
What is the median of the following data values? 108,102,43,100,22
PolarNik [594]

<u>Answer:</u> 100

<u>Reasoning:</u> The median is the "middle value" and to find it put the numbers in order from least to greatest.

22, 43, 100, 102, 108.

now take a number from both sides

43, 100, 102

Once again.

100

So 100 is the median.

6 0
3 years ago
Read 2 more answers
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