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Kryger [21]
2 years ago
15

Given circle with radius of 5cm. Find out (a) diameter (b) perimeter (c) area

Mathematics
1 answer:
lidiya [134]2 years ago
8 0

Answer:

(1) 10

(2) 10π

(3) 25π

Step-by-step explanation:

We are given the radius of the circle which is halfway through a circle. The radius of this circle is 5.

We are required to find the diameter, perimeter, and area using the given data we have been provided with.

Let's start:

#1: Diameter

To find diameter d use formula:

                                               d=2*r

After substituting  r=5 we have:

                                             d=2*5\\d=10

After calculating, we have found that the diameter of the circle is 10.

#2: Perimeter

The perimeter can also be known as the circumference of a circle as they both measure the outline.

To find circumference C use formula:

                                                  C=2*r*\pi

After substituting r=5 we have:

                                          C=2*5*\pi

                                           C=10\pi

Convert 10\pi to a decimal and round it to the nearest tenth and we have:
⇒ 31.4159265359

⇒ 31.4

After calculating, we have found that the perimeter (or circumference of the circle is 10π or 31.4 (Rounded to the nearest tenth in decimal form).

#3: Area

To find area A use the formula:

                                            A=r^2*\pi

After substituting  r=5 we have:

                                           A=5^2*\pi \\A=25*\pi \\A=25\pi

Convert 25\pi to a decimal and round it to the nearest tenth and we have:

⇒ 78.5398163397

⇒ 78.6

After calculating we have found that the area of the circle is either 25π or 78.6 (rounded to the nearest tenth).

After finding all the required information, we have now completed the answer.

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malfutka [58]

Given that, the variable z is directly proportional to x.

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Consider the function f(x)=9-x^2/x^2-4 For which intervals is f(x) positive? Check ALL that apply.
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Answer:

a. f (x) < 0 for x ∈ (-∞ ,-3)

b. f (x) > 0 for x ∈ (-3,-2)

c. f (x) < 0 for x ∈ (-2,2)

d. f (x) > 0 for x ∈ (2,3)

e.f (x) < 0 for x ∈ (3,∞)

Step-by-step explanation:

Here, the given function is:f(x)=   \frac{9-x^2}{x^2-4}

Now, to check for the sign of f(x) at x = k, put the value of x from the given interval.

We get:

<u>a. (-infinity, -3) </u>

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⇒ f (x) < 0 for x ∈ (-∞ ,-3)

b. (-3, -2)

put k = -2.5 from the given interval

We get f(-2.5)=   \frac{9-(-2.5)^2}{(-2.5)^2-4}  = \frac{9-6.25}{6.25-4}  = \frac{2.75}{2.25}  > 0

⇒ f (x) > 0 for x ∈ (-3,-2)

c. (-2, 2)

put k = 0 from the given interval

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d. (2, 3)

put k =2.5 from the given interval

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e. (infinity, 3)

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