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irina1246 [14]
3 years ago
14

Use the strip diagram to express 103 187 426

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
4 0
Need a picture to answer
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mrs long drive from nashville tennessee to atlanta georgia at an average rate of 50 miles per hour. The following equation can b
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Which statement follows from the given theorem? Theorem: The diagonal of a parallelogram divides it into two congruent triangles
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Answer with explanation:

Theorem : The diagonal of a parallelogram divides it into two congruent triangles.

Properties of parallelogram

1. Opposite sides are equal and parallel.

2. Diagonal bisect each other.

3. Opposite Angles are equal.

If you will look at the first three options

A:  Diagonals of a parallelogram bisect opposite angles.→→False(Alternate angles are equal)

B: Consecutive sides of a parallelogram are congruent.→→→False(Opposite sides are congruent)

C: Consecutive angles of a parallelogram are congruent.→→→False(Opposite angles are congruent)

D: Consecutive angles of a parallelogram are supplementary.→→→→True

For,any parallelogram, having vertices , A, B and C and D

Opposite sides are parallel, so

1.∠A + ∠B=180°

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Option D

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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
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