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Dahasolnce [82]
3 years ago
8

Hi what is the answer for this

Mathematics
2 answers:
TiliK225 [7]3 years ago
5 0

Answer:

formula = pi× r²× 1/4

Step-by-step explanation:

since radius = 8

3.14×8×8 = 3.14×16

4

therefore the answer is pi × 16

valentina_108 [34]3 years ago
3 0

Answer:

16\pi

Step-by-step explanation:

area of a circle is pi x r x r

since it is a quarter

1/4 x pi x 8 x 8 = 16 x pi

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How does lack of sleep affect risk of injury?
prohojiy [21]

Answer:

D. It interferes with motor responses

Step-by-step explanation:

Don't ask why I know this

8 0
3 years ago
Mark had $436. chris has $86 less than mark.write an expression to show how much money chris actually had
Nikitich [7]

We have been given that Mark had $436.

Further, we are given that Chris has $86 lesser than what mark had.

Let us assume that Chris has $x. Therefore, we can set up an equation:

(Amount Chris had) + $86 = (Amount Mark had)

x+86=436

Upon subtracting 86 from both the sides, we get:

x=436-86

Therefore, the required expression that represent the amount of money that Chris had is 436-86.


7 0
3 years ago
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
3 years ago
What key features can be identified from graphs of polynomials of higher degrees?
devlian [24]

Polynomials are algebraic expressions that contain more than two terms. AN example would be: f(x) = x^3 + x^2 + x +1. This equation contains three terms, with the 3rd degree as its highest term. It also means that the graph passed three x-intercepts. This depends in the highest degree. So, the first thing you do is plot the intercepts because for sure, the graph will pass there.

Hope this helps :)

4 0
4 years ago
What would be the expression after distributing, 3(x+5)?
Inessa05 [86]

Answer:

3x+15

Step-by-step explanation:

4 0
4 years ago
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