All categories

2 years ago

If the radius of the Cicular ground is 6mfind, the area of ground #plz help

Mathematics

1 answer:

Fed [463]2 years ago

3 0

Answer:

113.142857 m²

Step-by-step explanation:

As per the provide information in the given question, we have :

Radius of the circular ground = 6 m

We are asked to find the area of the ground.

Since, it is in the shape of circle, so we'll apply here the formula to find the area of the circular ground.

We know that,

>> Area of circle = πr

>> Area of the ground = $\sf \dfrac{22}{7}$ × 6 m × 6 m

>> Area of the ground = $\sf \dfrac{22}{7}$ × 36 m²

>> Area of the ground = $\sf \dfrac{792}{7}$ m²

>> Area of the ground = 113.142857 m²

$\therefore$ Therefore, area of the ground is 113.142857 m². 

You might be interested in

Create a Sample Space diagram and work out the probability - There is a 4 sided dice (numbers 1 to 4) and an 8 sided dice (numbe

tester [92]

Answer:

7 / 16

Step-by-step explanation:

Sample space is attached below :

Theoretical Probability of any event A

P(A) = (number of required outcome / Total number of possible outcomes)

Required outcome = sum greater than 7 = 14

Total number of possible outcomes = 32

P(sum greater than 7) = 14 / 32

P(sum greater than 7) = 7 / 16

3 0

2 years ago

Jamal makes $186 a week at his job. He plans to save 5% of his paycheck for a P.Y.F account. How much money will Jamal put in hi

kenny6666 [7]

Answer:

$9.30

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Suppose that F(x) = x2 and G(X) = 2/3 x^2. Which statement best compares the graph of G(x) with the graph of F(x)?

masha68 [24]

<h2>Answer:The graph of $G(x)$ is the graph of $F(x)$ compressed vertically.</h2>

Step-by-step explanation:

Given that $F(x)=x^{2}$ and $G(x)=\frac{2}{3}x^{2}$

$F(x)$ is always positive because $x^{2}$ is always positive.

$G(x)$ is always positive because $\frac{2}{3}x^{2}$ is always positive.

So,both are always positive.

So,there is no flipping over x-axis.

In $F(x)$ ,the height of a point at $x_{0}$ is $x_{0}^{2}$

In $G(x)$ ,the height of a point at $x_{0}$ is $\frac{2}{3}x_{0}^{2}$

So,height of any point has less height in $G(x)$ than $F(x)$

So,the graph of $G(x)$ is the graph of $F(x)$ compressed vertically.

3 0

2 years ago

If 400 bricks measuring 81" by 31" are required to build a wall 42 ft. high, how long is the wall

musickatia [10]

Idek lol but yeah have a good day

7 0

2 years ago

Given: triangle ABC and triangle EDC, C is the midpoint of BD and AE. Prove: AB || DE

VladimirAG [237]

Step-by-step explanation:

1. ΔABC=ΔCDE: a) according to the condition BC=CD and AC=CE; b) m∠(BCA)=m∠(DCE).

2. if ΔABC=ΔCDE, then m∠(BAC)=m∠(DEC) and m∠(ABC)=m∠(CDE).

3. if m∠(BAC)=m∠(DEC), then AB || DE (or if m∠(ABC)=m∠(CDE), then AB || DE).

5 0

2 years ago

If the radius of the Cicular ground is 6mfind, the area of ground #plz help​

If the radius of the Cicular ground is 6mfind, the area of ground #plz help