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3 years ago

Can someone help me asap

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

8 0

Answer:

the top one would be my best he is because it has the two sybles in the problem they gave you

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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:

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

Evaluate for dy/dt:

$\displaystyle \frac{dy}{dt} = -\frac{(15)(2)}{(36)}=-\frac{5}{6}$

Finally, using dA/dt, substitute in appropriate values:

$\displaystyle \frac{dA}{dt} = \frac{1}{2}\left((36)(2)+(15)\left(-\frac{5}{6}\right)\right)$

Evaluate. Hence:

$\displaystyle \frac{dA}{dt} = \frac{119\text{ ft}^2}{4\text{ s}} = 29.75\text{ ft$^2$/s}$

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

8 0

3 years ago

It takes 32 minutes for 8 people to paint 8 walls. How many minutes does it take 12 people to paint 12 walls ? Confused !

Radda [10]

Answer:

It takes 32 Minutes for 12 people to paint 12 walls

Step-by-step explanation:

If it takes 8 people 32 minutes to paint 8 walls, then that means each person takes 32 minutes to paint 1 wall. With that being said, If you have 12 people painting 12 walls, It will take 32 minutes for all 12 walls to be painted.

Hope this helps!

4 0

3 years ago

Read 2 more answers

Which represents the solution(s) of the system of equations, y = -x2 + 6x + 16 and y = - 4x + 37? Determine the solution set

AysviL [449]

Answer:

The answer is c

Step-by-step explanation:

(7,9) (3,25)

3 0

3 years ago

Si un proyectil asciende verticalmente, y después de 3 segundos alcanza su altura máxima, calcule la velocidad que lleva a la mi

lys-0071 [83]

Answer:

The speed is 20.8 m/s

Step-by-step explanation:

If a projectile ascends vertically, and after 3 seconds it reaches its maximum height, calculate the velocity that it carries to the middle of its downward trajectory

Let the maximum height is h and initial velocity is u.

From first equation of motion

v = u + at

0 = u - g x 3

u = 3 g.....(1)

Use third equation of motion

$v^2 = u^2 - 2 gh \\\\0 = 9 g^2 - 2 gh \\\\h = 4.5 g$

Let the speed at half the height is v'.

$v^2 = u^2 + 2 gh \\\\v'^2 = 0 + 2 g\times 2.25 g\\\\v'^2 = 4.5\times 9.8\times9.8\\\\v' = 20.8 m/s$

8 0

3 years ago

Solve for X. Help please!

nataly862011 [7]

Answer:

Step-by-step explanation:

180-80=100

100+5x+10+6x+4=180

114+11x=180

11x=66

x=6

4 0

3 years ago

Can someone help me asap​

Can someone help me asap