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2 years ago

Find the radius of a circe given the area 25pi square inch.

Mathematics

1 answer:

Elis [28]2 years ago

6 0

Answer:

Radius = 5 inches

Step-by-step explanation:

1) We have to use the equation for the area of a circle to find the radius of the circle.

Area of a circle = πr^2

2) Plug in the information we already know, i.e., the area.

25π = πr^2

3) Cancel out the π on each side.

25 = r^2

4) Take the square root of 25

r = $\sqrt{25\\}$

r = 5

5) The radius is 5 inches.

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Line s has an equation of y=1/3-X-5. Line t is perpendicular to line s and passes through (-2, 9). What is the equation of line

anyanavicka [17]

Answer:

y = 3x + 15

Step-by-step explanation:

1. Find the reciprocal of the slope and change signs: -3

2. y = 3x + b

3. Plug in x and y --> b = 15

4. Make your equation using y = mx + b: y = 3x + 15

6 0

1 year ago

Read 2 more answers

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta = \dfrac{1}{9} \tan \theta+\text{C}$

$\textsf{Use the trigonometric identity}: \quad \tan \theta=\dfrac{\sin \theta}{\cos \theta}$

$\implies \dfrac{\sin \theta}{9 \cos \theta} +\text{C}$

$\textsf{Substitute back in } \sin \theta=\dfrac{x}{3}:$

$\implies \dfrac{x}{9(3 \cos \theta)} +\text{C}$

$\textsf{Substitute back in }3 \cos \theta=\sqrt{9-x^2}:$

$\implies \dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Learn more about integration by substitution here:

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4 0

2 years ago

The difference of 2 and the product of 3and k

allsm [11]

2-3k because the difference between something means it is subtraction and the product of 3 and k would just be 3k since you don't know what k is.

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3 years ago

Where is the axis on a quadrant

dezoksy [38]

Answer:

x axis and y axis divide the coordinate plane into four quadrants.

Step-by-step explanation:

If one coordinate is zero, then the point is located on the x-axis or the y-axis. If y coordinate is zero, then the point is located on the y-axis. If both coordinates are zero, then the point represents the origin.

8 0

3 years ago

Find the missing sides, and give the simplified form of the square root

Brilliant_brown [7]

Answer:

in the pics

Step-by-step explanation:

In case you are wondering, I used the pythagorean theorem. If you want to learn how to use it I suggest you google it, there are plenty of sites that explain how to use it.

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below :)

7 0

3 years ago