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irga5000 [103]
2 years ago
12

What is x - 5= 49 - 14(x+6)

Mathematics
1 answer:
PilotLPTM [1.2K]2 years ago
8 0

Answer:

x = -2

Step-by-step explanation:

First, distribute the -14(x+6).

x-5 = 49-14(x+6)

x-5 = 49-14x-84

Next, add up all the like terms (49+(-84))

x-5 = 49-14x-84

x-5 = -14x-35

Now, bring all the like terms to one side. All the x's to one side and all the numbers to the other side.

x-5 = -14x-35

x+14x = -35+5

15x = -30

Last, get x by itself.

15x = -30

x = -2

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A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2
Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

  • (a) The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

<u />

  • (b) The rate the area formed by the ladder is changing is approximately <u>-75.29 ft.²/sec</u>

<u />

  • (c) The rate at which the angle formed with the wall is changing is approximately <u>0.286 rad/sec</u>.

Reasons:

The given parameter are;

Length of the ladder, <em>l</em> = 25 feet

Rate at which the base of the ladder is pulled, \displaystyle \frac{dx}{dt} = 2 feet per second

(a) Let <em>y</em> represent the height of the ladder on the wall, by chain rule of differentiation, we have;

\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}

25² = x² + y²

y = √(25² - x²)

\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}

Which gives;

\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}\times \frac{dx}{dt} =  \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}\times2

\displaystyle \frac{dy}{dt} =  \mathbf{ \frac{x \cdot \sqrt{625-x^2}  }{x^2- 625}\times2}

When x = 15, we get;

\displaystyle \frac{dy}{dt} =   \frac{15 \times \sqrt{625-15^2}  }{15^2- 625}\times2 = \mathbf{-1.5}

The velocity of the top of the ladder = <u>1.5 m/s downwards</u>

When x = 20, we get;

\displaystyle \frac{dy}{dt} =   \frac{20 \times \sqrt{625-20^2}  }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6

The velocity of the top of the ladder = \underline{-2.\overline{6} \ m/s \ downwards}

When x = 24, we get;

\displaystyle \frac{dy}{dt} =   \frac{24 \times \sqrt{625-24^2}  }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}}  \approx -6.86

The velocity of the top of the ladder ≈ <u>-6.86 m/s downwards</u>

(b) \displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}

Therefore;

\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}

\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}

\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}

Therefore;

\displaystyle \frac{dA}{dt} =  \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2

When the ladder is 24 feet from the wall, we have;

x = 24

\displaystyle \frac{dA}{dt} =  \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}

The rate the area formed by the ladder is changing, \displaystyle \frac{dA}{dt} ≈ <u>-75.29 ft.²/sec</u>

(c) From trigonometric ratios, we have;

\displaystyle sin(\theta) = \frac{x}{25}

\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}

\displaystyle \frac{d \theta}{dt}  = \frac{d \theta}{dx} \times \frac{dx}{dt}

\displaystyle\frac{d \theta}{dx}  = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}

Which gives;

\displaystyle \frac{d \theta}{dt}  =  -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}

When x = 24 feet, we have;

\displaystyle \frac{d \theta}{dt} =  -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is \displaystyle \frac{d \theta}{dt} ≈ <u>0.286 rad/sec</u>

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0
2 years ago
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