May someone please help me with this

If the area of your garden is ( d^2 -16 ) and the length is (d+4

Alex17521 [72]

Answer:

d^3+4d^2-16d-64

Step-by-step explanation:

d^d+d^2*4+(-16)d+(-16)*4

d^2d+4d^2-16d-16*4

5 0

2 years ago

What expression represents the perimeter of the rectangle?<br> shown below

Komok [63]

Answer:

4x + 4

Step-by-step explanation:

perimeter= breadth + length + breadth + length

(x+2)+(x+2)+x+x

= 4x(because there is 4 x) + 2 +2

= 4x + 4

5 0

3 years ago

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 $\frac{dP}{dt}$

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

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

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

7 0

3 years ago

Help me pls thankyouuuuuuu

torisob [31]

Answer: 4 $\sqrt[3]{4}$

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

HELP PLEASE! Thank you!

Nataly [62]

Answer:

See attached graph for the first part

Answer to second part: The end part of the graph show the slowest increase

Step-by-step explanation:

The attached picture represents the number of infected people, starting with a relatively small number at the origin of the horizontal axis (x=0, or time=0) then increasing abruptly in the center of the graph with steep slope. and then infection slowing down (although still slowly increasing) in the region highlighted in yellow to the right of the graph.

8 0

3 years ago