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3 years ago

What is the perimeter of the quadrilateral that is the length of 7 height of 8

Mathematics

2 answers:

scoray [572]3 years ago

7 0

Answer:

<h2>30 units</h2>

Step-by-step explanation:

assuming it is a rectangle, there is no information on the shape of the quadrilateral, add all sides.

7 + 8 + 7 + 8 = 30 units

marta [7]3 years ago

6 0

Answer:

Step-by-step explanation:

What you need to do is add all the sides when you want to find perimeter so for this one you would want to do 8+8+7+7 and that should give you 30. I hope I could help!!

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Work out the value of each expression, when r= 2.5 and t = 6.

Usimov [2.4K]

Answer:

a) 4(t - r) = 4(6 - 2.5) = 4 x 3.5 = 14

b) 4t - r = (4 x 6) - 2.5 = 24 - 2.5 = 21.5

c) 2(3t - 10) = 2(3 x 6 - 10) = 2(18 - 10) = 2 x 8 = 16

d) (t - 2)² = (6 - 2)² = 4² = 4 x 4 = 16

e) cannot decipher the equation

f) 6r + t = 6 x 2.5 + 6 = 15 + 6 = 21

8 0

3 years ago

Read 2 more answers

A building has a height of 100ft and a length of 100ft. An architect wants to add to the building by doubling the length of the

skelet666 [1.2K]

That would be 200ft sir

5 0

4 years ago

Add. –32.7+(–0.2) a. –32.9 b. –32.5 c. 32.5 d. 32.9

yKpoI14uk [10]

The answer to your question is -32.9

5 0

3 years ago

At the end of a recent WNBA regular season, the Phoenix murmur had 12 more victories than losses. The number of victories they h

Hoochie [10]

Answer:

34 games

Step-by-step explanation:

let the number of losses be $l$

let the number of victories be $v$

the Phoenix murmur had 12 more victories than losses:

$v=l+12$

The number of victories they had was one more than two times the number of losses:

$v=2l+1$

We have 2 expressions for $v$ , equating these 2, we can solve for $l$ :

$l+12=2l+1\\12-1=2l-l\\11=l$

So 11 games, they lost.

Using this value, we can plug into the 1st equation to solve for v:

$v=l+12\\v=11+12\\v=23$

So 23 games, they won.

Given that no draws, they played a total of 23 + 11 = 34 games this season

6 0

4 years ago

Little Snail is going to visit his friend over at the next pond, 3 miles away. He can crawl ( 1/2. 1/3, 1/4, 3/4, 2/3 ) of a mil

KIM [24]

Solution :

It is given that Little Snail is going to visit one of his friend at the pond which is 3 miles away.

When the snail crawls 1/2 of a mile per day, it will take him, $1 \times \frac{2}{1} \times 3$

= 6 days to get to the next pond.

When the snail crawls 1/3 of a mile per day, it will take him, $1 \times \frac{3}{1} \times 3$

= 9 days to get to the next pond.

When the snail crawls 1/4 of a mile per day, it will take him, $1 \times \frac{4}{1} \times 3$

= 12 days to get to the next pond.

When the snail crawls 3/4 of a mile per day, it will take him, $1 \times \frac{4}{3} \times 3$

= 4 days to get to the next pond.

When the snail crawls 2/3 of a mile per day, it will take him, $1 \times \frac{3}{2} \times 3$

= 4.5 days to get to the next pond.

3 0

3 years ago