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3 years ago

If f(x) = –4x2 – 6x – 1 and g(x) = –x2 – 5x + 3, find (f – g)(x).

Mathematics

2 answers:

nexus9112 [7]3 years ago

8 0

Answer:

(f – g)(x) = -3x² - x -4.

Step-by-step explanation:

Given : If f(x) = –4x² – 6x – 1 and g(x) = –x²– 5x + 3.

To find : (f – g)(x).

Solution ; We have given that f(x) = –4x² – 6x – 1

g(x) = –x²– 5x + 3.

(f – g)(x) = f(x) - g(x).

(f – g)(x) = (–4x² – 6x – 1 ) - ( –x²– 5x + 3) .

(f – g)(x) = –4x² – 6x – 1 + x²+ 5x - 3 .

Combine like terms

(f – g)(x) = –4x² + x² - 6x + 5x -1 -3.

(f – g)(x) = -3x² - x -4.

Therefore, (f – g)(x) = -3x² - x -4.

Leokris [45]3 years ago

5 0

(-4x^2 -6x -1) - (-x^2 -5x +3) = -3x^2 - x - 4

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8 0

3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows