Record the lengths of the sides of ABC and ADE.

An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a ra

lukranit [14]

Answer:

$\frac{dA}{dt} = 28800 \ m^2/year$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Geometry

Area of a Rectangle: A = lw

Algebra I

Exponential Property: $w^n \cdot w^m = w^{n + m}$

Calculus

Derivatives

Differentiating with respect to time

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

Area is A = lw

2w = l

w = 300 m

$\frac{dw}{dt} = 24 \ m/year$

Step 2: Rewrite Equation

Substitute in l: A = (2w)w
Multiply: A = 2w²

Step 3: Differentiate

Differentiate the new area formula with respect to time.

Differentiate [Basic Power Rule]: $\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}$
Simplify: $\frac{dA}{dt} = 4w\frac{dw}{dt}$

Step 4: Find Rate

Use defined variables

Substitute: $\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = (1200 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = 28800 \ m^2/year$

3 0

3 years ago

Read 2 more answers

Derive a general formula for projectile motion with the object launched at an angle

vagabundo [1.1K]

Answer:

y = x tan θ - (g / 2v₀² cos² θ) x²

Explanation:

An equation is called a general formula that relates the position on the x-axis and the height on the body's axis.

Let's write the position on each axis

X axis. No acceleration

x = v₀ₓ t

Y Axis. There is the acceleration of gravity

y = $v_{oy}$ t - ½ g t²

Let's clear the time in the first equation and substitute in the second

y = v₀ sin θ (x / v₀ₓ) - ½ g (x / v₀ₓ)²

y = v₀ sin θ / v₀ cos θ x - ½ g x² / v₀² cos² θ

y = x tan θ - (g / 2v₀² cos² θ) x²

This is the trajectory equation in projectile launching

6 0

3 years ago

Why can water dissolve some ionic compounds, like NH4Cl, as well as some nonionic compounds, like methanol?

nirvana33 [79]

I think that’s the answer u r looking for

4 0

3 years ago

Read 2 more answers

Which has more momentum, a 2.0 kg dog at 41 m/s or a 75 kg pony at 1.0 m/s?

Nesterboy [21]

Answer:

The dog has more momentum than the pony.

Explanation:

To solve this problem, we must remember the formula for calculating momentum, which is given below:

momentum = p = m*v

where m represents the mass of the object and v represents the velocity of the object

Using this knowledge, let's calculate the momentum for the dog and the pony.

Dog: p = m*v = (2kg)*(41 m/s) = 82 kg*m/s

Pony: p = m*v = (75kg)*(1 m/s) = 75 kg*m/s

Since 82 > 75, we can conclude that the dog has more momentum.

Hope this helps!

3 0

4 years ago

In physics what do we use machines for

Tasya [4]

Answer:

hope it helps for you

Explanation:

A machine is an object or mechanical device that receives an input amount of work and transfers the energy to an output amount of work. For an ideal machine, the input work and output work are always the same. The six common simple machines are the lever, wheel and axle, pulley, inclined plane, wedge, and screw.

7 0

3 years ago