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anastassius [24]
3 years ago
15

The length of a rectangle is 4 cm less than 3 times its width. If the width is x cm, then the length is

Mathematics
1 answer:
Feliz [49]3 years ago
6 0

Answer:

Length: 3w-4

Step-by-step explanation:

Length: 3w-4

Width: W

<em>*You don't have to, but I am currently trying to reach the next level, and all I need is some more brainliest answers. If you think my answer was brainly enough, you can make my answer the brainliest, but no pressure. I just help people for fun! :) Thank you, have a great day!*</em>

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What 3.405 ounces in expanded form
Y_Kistochka [10]
What you’re doing is breaking it down into its parts and writing it into an equation that way long hand.

3 + .4 + .00 + .005
7 0
3 years ago
Show how to determine the simplest form of by first finding the greatest common factor of 24 and 60
natima [27]

Answer:

The  greatest common factor would be 12.

Step-by-step explanation:

24: 1, 2, 3, 4, 6, 8, <u><em>12</em></u>, 24

60: 1, 2, 3, 4, 5, 6, 10 , <u><em>12</em></u>, 15, 20, 30, 60

Hope it helps!

4 0
3 years ago
What is the area of this figure? Enter your answer in the box.
Basile [38]
Luckily for us, the diagram already divided this figure into separate polygons. What I will be explaining is basically the addition of the areas of all the separate polygons. The area of the uppermost triangle is:

1/2 x b x h
= 1/2 x 20 x 8
(the base is 20, because in a parallelogram, opposite sides are congruent)
=10 x 8
= 80 in. squared

The next polygon we will be taking the area of is the parallelogram with the base length of 20 and the height of 16.

Area = b x h
= 20 x 16
= 320 in. squared

Now all we have left to do is add the two areas to obtain the total area.

Total Area = 320 + 80 = 400 in. squared


3 0
3 years ago
Jacob weights 36.4 pounds and alice. weights 42.275 pounds. how much do they weigh together
QveST [7]

Answer:

78.675 pounds

Step-by-step explanation:

36.4 + 42.275 = 78.675 pounds

5 0
3 years ago
Read 2 more answers
A group of 100 people is divided into 2 teams with 45 people in team A and 55 people in team B.
Zepler [3.9K]
Team A) 45 people
Team B) 55 people

A)There are two ways to solve this problem, finding the number of combinations possible for Team B, or the number of combinations possible for Team A. 
Team A
It's a given that 20 mathematicians are on team A, which leavs the other 25 people for team A to be chosen from a pool of 80 (100- 20 mathletes)
80-C-25 = 80! / (25!/(80-25)!) =<span>363,413,731,121,503,794,368
</span>or 3.63 x 10^20
Solving using Team B
Same concept, but choosing 55 from a pool of 80 (mathletes excluded)
80-C-25 = 80! / (55!(80-55!) = 363,413,731,121,503,794,368
or 3.63 x 10^20

As you can, we get the same answer for both. 

B)
If none of the mathematicians are on team A, then we exclude the 20 and choose 45:
80-C-45 = 80! / (45!(80-45)!) = <span>5,790,061,984,745,3606,481,440
or 5.79 x 10^22

Note that, if you solve from the perspective of Team B (80-C-35), you get the same answer</span>
7 0
3 years ago
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