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2 years ago

8

A water sprinkler sends water out in a circular pattern. How large is the watered area if the radius of the watering pattern is2

3ft ft?

1 answer:

Mamont248 [21]2 years ago

6 0

Answer:

Step-by-step explanation:

Simply put, it is similar to finding the area of a circle.

Therefore, a=pi*r^2

a=pi*23^2

a=529pi

a=1661.9 ft^2

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Abc = dfe find the value of x.

LiRa [457]

BC <span>≅ EF (CPCTC)

Set the two side lengths equal to each other.

38.4 = 5x - 10.6

Isolate the x. Add 10.6 to both sides

38.4 (+10.6) = 5x - 10.6 (+10.6)

5x = 49

Isolate the x, divide 5 from both sides

5x/5 = 49/5

x = 49/5

x = 9.8

9.8 should be your answer

hope this helps</span>

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3 years ago

Read 2 more answers

Differentiating Functions of Other Bases In Exercise, find the derivative of the function.

aniked [119]

Answer:

$\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}$

Step-by-step explanation:

We are given the following in the question:

$y = \ln {16}(x^2 - 3x)$

We have to find the derivative of the given expression.

$y = \ln 16(x^2 - 3x)\\\text{Using the log propert}\\\\\log_a b = \dfrac{\log b}{\log a}\\\\dfrac{d(x^n)}{dx} = nx^{n-1}\\\\\dfrac{d(\log x)}{dx} = \dfrac{1}{x}\\\\\text{\bold{Differentiating we get}}\\\\\displaystyle\frac{dy}{dx} = \frac{d(\ln 16(x^2-3x))}{dx}\\\\= \frac{1}{16(x^2-3x)})\frac{d(x^2-3x)}{dx}\\\\=\frac{1}{\log 16(x^2-3x)}(2x - 3)\\\\= \frac{2x-3}{\ln 16(x^2 - 3x)}$

$\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}$

7 0

4 years ago

Is the function y=10 linear or nonlinear?

Alja [10]

Nonlinear because its not changing in y values

4 0

3 years ago

Select the correct answer.

EleoNora [17]

B? I think ( do not come at me if it’s wrong)

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3 years ago

3. The data set shows the number of practice throws players in a basketball competition made and the number of free throws they

babymother [125]

<u>Answer-</u>

<em>The </em><em>exponential model</em><em> best fits the data set.</em>

<u>Solution-</u>

x = input variable = number of practice throws

y = output variable = number of free throws

Using Excel, Linear, Quadratic and Exponential regression model were generated.

The best fit equation and co-efficient of determination R² are as follows,

<u>Linear Regression</u>

$y=2.155x+0.391,\ R^2=0.903$

<u>Quadratic Regression</u>

$y=0.096x^2+0.713x+3.803,\ R^2=0.948$

<u>Exponential Regression</u>

$y=4.625e^{0.141x},\ R^2=0.951$

The value of co-efficient of determination R² ranges from 0 to 1, the more closer its value to 1 the better the regression model is.

Now,

$R^2_{Linear}< R^2_{Quadratic}< R^2_{Exponential}$

Therefore, the Exponential Regression model must be followed.

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4 years ago