PLEASE I NEED HELP ASAP!!!!

Mathematics

2 answers:

Svetllana [295]3 years ago

6 0

The vertex form is
$y = a {(x - h)}^{2} + k$
where a is the coefficient, and (h,k) is the vertex. Looking at the graph, we see that the vertex is (-2,2), so h is -2 and k is 2. We see that the parabola opens upwards, so A is positive, and is stretched by a factor of 3. (We can find this by looking where the (1,1) should be. We go over 1 from the parabola, and see where the y coordinate is. This is at (-1,5), or 1 unit right and 3 units up.) Therefore, our a is 3. Plugging this into the vertex form:
$y = 3 {(x + 2)}^{2} + 2$
Which is answer A

Black_prince [1.1K]3 years ago

6 0

Filling this spot in and agree that the answer is A.

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Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$