you plant a spruce tree that grows 4 inches per year and a hemlock tree that grows 6 inches per year.

Mathematics

1 answer:

gizmo_the_mogwai [7]2 years ago

4 0

Answer:

this doesn't even make sense.

Step-by-step explanation:

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a driver was fined $20.00 plus $4.00 for every mile per hour over the speed limit of 55 mph. is the fine was $76, how fast was t

Ksenya-84 [330]

20+4x=76 subtract 20 from both sides

4x=56 divide both sides by 4

x=14

So he was driving at 55+14=69 mph

3 0

3 years ago

What are the determinants for solving this linear system?<br> 5x + 3y = 17<br> −8x − 3y = 9

Ksivusya [100]

For this case we have the following system of equations:
5x + 3y = 17
-8x - 3y = 9
We can rewrite the system like:
Ax = b
Where,
A = [5 3; -8 -3]
b = [17; 9]
x = [x; y]
The determinant of matrix A is given by:
lAl = ((5) * (- 3)) - ((3) * (- 8))
lAl = (-15) - (-24)
lAl = -15 + 24
lAl = 9
Answer:
The determinant for solving this linear system is:
lAl = 9

3 0

2 years ago

Read 2 more answers

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

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 $\frac{dP}{dt}$

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