Can you guys help me with all 3 questions please would really appreciate it.

Mathematics

1 answer:

Temka [501]3 years ago

3 0

Answer:

Number 1 is 17 cents

Number 2 is 164.35

Number 3 is 213

Step-by-step explanation:

Please mark brainliest

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3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

LenaWriter [7]

Answer:

The area of the triangle formed is increasing at a rate of 29.75 square feet per second.

Step-by-step explanation:

A 39-foot ladder is leaning against a vertical wall. We are given that the bottom of the ladder is being pulled away at a rate of two feet per second, and we want to find the rate at which the area of the triangle being formed is is changing when the bottom of the ladder is 15 feet from the wall.

Please refer to the diagram below. x is the distance from the bottom of the ladder to the wall and y is the height of the ladder on the wall.

According to the Pythagorean Theorem:

$\displaystyle x^2+y^2=1521$

Let's take the derivative of both sides with respect to time t. Hence:

$\displaystyle \frac{d}{dt}\left[x^2+y^2\right] = \frac{d}{dt}\left[ 1521\right]$

Implicitly differentiate:

$\displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

Simplify:

$\displaystyle x\frac{dx}{dt} + y \frac{dy}{dt} = 0$

The area of the triangle formed will be given by:

$\displaystyle A = \frac{1}{2} xy$

Again, let's take the derivative of both sides with respect to time t:

$\displaystyle \frac{dA}{dt} = \frac{d}{dt}\left[\frac{1}{2}xy\right]$

From the Product Rule:

$\displaystyle \frac{dA}{dt} = \frac{1}{2}\left(y\frac{dx}{dt} + x\frac{dy}{dt}\right)$

At that instant, the ladder is 15 feet from the base of the wall. So, x = 15. Using this information, find y.

$\displaystyle y = \sqrt{1521-(15)^2}=36$

The bottom of the ladder is being pulled away from the wall at a rate of two feet per second. So, dx/dt = 2. Using this information and the first equation, find dy/dt:

$\displaystyle \frac{dy}{dt} =-\frac{x\dfrac{dx}{dt}}{y}$