All categories

3 years ago

The area of a rectangular field is 200 m²and its perimeter is 60 m. Find the lengthand width of the field.

Mathematics

2 answers:

attashe74 [19]3 years ago

8 0

Answer:

length: 20m

width: 10m

(or vice versa)

Step-by-step explanation:

By substituting each of the values (20 meters for length and 10 meters for width), you will receive answers that satisfy each of the given traits of the rectangle:

200=20x10

200=200

60=2(20)+2(10)

60=40+20

60=60

Mademuasel [1]3 years ago

7 0

The length and width are 10 and 20

You might be interested in

What's 7 hours - 30mins ?

Naddik [55]

Answer:

390 mins (or) 6 hours 30 mins

Step-by-step explanation:

7 hours = 420 mins

420 mins - 30 mins = 390 mins

390 mins = 6 hours 30 mins

Have a nice day!

8 0

3 years ago

1) Determine if the side lengths given form a triangle

Margaret [11]

Using the triangle inequality theorem, we can figure this out.

2- Yes

3- No

4- No

5- Yes

6- Yes

6 0

3 years ago

HELP ASAP (40 POINTS)

frez [133]

Answer:

c i think because the problem =-

$- \frac{1}{4}$

4 0

2 years ago

An organization has members who possess IQs in the top 4% of the population. If IQs are normally distributed, with a mean of 100

murzikaleks [220]

Answer:

The minimum IQ score will be "126".

Step-by-step explanation:

The given values are:

Mean

$\mu = 100$

Standard deviation

$\sigma=15$

Now,

⇒ $P(z>x)=4 \ percent \ i.e., 0.04$

⇒ $P(z>\frac{x- \mu}{\sigma} )=0.04$

⇒ $1-P(z \leq \frac{x-100}{15})=0.04$

⇒ $\frac{x-100}{15}=z0.96$

$=NORMDIS(z=0.96)$

$=1.751$

⇒ $x=100+15\times 1.751$

$=126.265 \ i.e., 126$

7 0

3 years ago

12.5% of what number 24?

Valentin [98]

Answer:

192

Step-by-step explanation:

12.5% is 24.

So 100% will be (100/12.5) x 24 = 192.

Done.

6 0

3 years ago

Read 2 more answers

The area of a rectangular field is 200 m²and its perimeter is 60 m. Find the lengthand width of the field.​

The area of a rectangular field is 200 m²and its perimeter is 60 m. Find the lengthand width of the field.