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Dvinal [7]
2 years ago
9

The radius of a circular merry-go-round is 3 meters. which measurement is closest to the area of the merry-go-round in square me

ters?
A. 18.84 m2
B. 37.68 m2
C. 113.04 m2
D. 28.26 m2​
Mathematics
1 answer:
djyliett [7]2 years ago
4 0

Answer:

area of the merry go round =πr²

=22/7×3²

=28.28

answer is option D

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Answer:

3235

Step-by-step explanation:

100 for the week plus 3135 which is 5.5% of 57,000

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What is the directrix of the parabola? (y+4)2=4(x−2)
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the answer is 1

Step-by-step explanation:

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3.102 x 10^3 in standard form
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3102x

Step-by-step explanation:

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3 years ago
Find the vertical and horizontal asymptotes, domain, range, and roots of f (x) = -1 / x-3 +2.
gladu [14]

Answer:

Vertical asymptote: x=3

Horizontal asymptote: f(x) =2

Domain of f(x) is all real numbers except 3.

Range of f(x) is all real numbers except 2.

Step-by-step explanation:

Given:

Function:

f (x) = -\dfrac{1 }{ x-3} +2

One root, x = 3.5

To find:

Vertical and horizontal asymptote, domain, range and roots of f(x).

Solution:

First of all, let us find the roots of f(x).

<em>Roots of f(x) means the value of x where f(x) = 0</em>

0= -\dfrac{1 }{ x-3} +2\\\Rightarrow 2= \dfrac{1 }{ x-3}\\\Rightarrow 2x-2 \times 3=1\\\Rightarrow 2x=7\\\Rightarrow x = 3.5

One root, x = 3.5

Domain of f(x) i.e. the values that we give as input to the function and there is a value of f(x) defined for it.

For x = 3, the value of f(x) \rightarrow \infty

For all, other values of x , f(x) is defined.

Hence, Domain of f(x) is all real numbers except 3.

Range of f(x) i.e. the values that are possible output of the function.

f(x) = 2 is not possible in this case because something is subtracted from 2. That something is \frac{1}{x-3}.

Hence, Range of f(x) is all real numbers except 2.

Vertical Asymptote is the value of x, where value of f(x) \rightarrow \infty.

-\dfrac{1 }{ x-3} +2 \rightarrow \infty

It is possible only when

x-3=0\\\Rightarrow x=3

\therefore vertical asymptote: x=3

Horizontal Asymptote is the value of f(x) , where value of x \rightarrow \infty.

x\rightarrow \infty \Rightarrow \dfrac{1 }{ x-3} \rightarrow 0\\\therefore f(x) =-0+2 \\\Rightarrow f(x) =2

\therefore Horizontal asymptote: f(x) =2

Please refer to the graph of given function as shown in the attached image.

5 0
3 years ago
Sqr root 8+3 Sqr root 2+ Sqr root 32 simplest form
Kisachek [45]

Answer:

<u>9\sqrt{2} </u>

Step-by-step explanation:

The given question is :

\sqrt{8} +3\sqrt{2} +\sqrt{32}

8 = 4 * 2

32 = 16 * 2

\sqrt{8} =\sqrt{4*2} = 2\sqrt{2} \\\sqrt{32} =\sqrt{16*2} =4\sqrt{2}

∴ \sqrt{8} +3\sqrt{2} +\sqrt{32}

= 2\sqrt{2} +3\sqrt{2} +4\sqrt{2}

= 9\sqrt{2}

6 0
3 years ago
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