All categories

3 years ago

Find the perimeter of a regular octagon whose side is "a" units

Mathematics

1 answer:

Zinaida [17]3 years ago

4 0

Answer:

I think its 8a because in a regular octagon we have 8 sides.

You might be interested in

1 2/3 times 3 2/3 greater than 3 2/3

jek_recluse [69]

Yes, yes it is :)
your welcome

3 0

3 years ago

Michael plays on his school's basketball team. Last year, he made 98% of his baskets. He attempted 200 shots. How is 98% express

slamgirl [31]

98% expressed as a fraction

= 98/100

Option C

3 0

1 year ago

A certain animated movie earned $1.1*10^9 in revenues at the box office. The movie lasts 9.1*10^1 minutes. How much revenue was

Vadim26 [7]

Answer:

$12,087,912.1

Step-by-step explanation:

$1.1*10^9$ is in scientific notation, to get this into normal number, we take the decimal point and move it "9" times to the right [whatever power of 10 we have, we move to right (if positive).] Then we fillup the rest with zeros.

So we have:

$1.1*10^9=1,100,000,000$

This is the revenue.

We do similar process for time:

$9.1*10^1=91$

This is the time , in minutes.

We want revenue PER minute, that means we divide the total revenue by total time, that would be:

$\frac{1,100,000,000}{91}=12,087,912.1$

The revenue earned per minute is $12,087,912.1

6 0

3 years ago

Math-- Farmer Bill has 500 meters of fencing and wants to enclose a rectangular plot that borders on a river. If farmer Bill doe

bixtya [17]

Answer:

Required dimensions of the rectangle are L = 200 m, W = 100 m

The largest area that can be enclosed is 20,000 sq m.

Step-by-step explanation:

The available length of the fencing = 500 m

Now, Perimeter of a rectangle = SUM OF ALL SIDES = 2(L+B)

But, here once side of the rectangle is NOT FENCED.

So, the required perimeter

= Perimeter of Complete field - Boundary of 1 open side

= 2(L+ W) - L = 2W + L

Now, fencing is given as 500 m

⇒ 2W + L = 500

Now, to maximize the length and width:

put L = 200, W = 100

we get 2(W) +L = 2(200) + 100 = 500 m

Hence, required dimensions of the rectangle are L = 200 m, W = 100 m

The maximized area = Length x Width

= 200 m x 100 m = 20, 000 sq m

Hence, the largest area that can be enclosed is 20,000 sq m.

6 0

3 years ago

At lift-off, a rocket weighs 40.2 tons, including the weight of fuel. It is fired vertically, and the fuel is consumed at the ra

Thepotemich [5.8K]

Step-by-step explanation:

Below is an attachment containing the solution.

8 0

3 years ago