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coldgirl [10]
3 years ago
6

How to write a real world problem that could be solved using proportions

Mathematics
2 answers:
julsineya [31]3 years ago
6 0
You could used building measurements, like the building is 85 ft tall and 45 ft wide, and the second build is 67 ft tall, find the width of the second building. Set up a proportion with the first building measurements and the second building measurements using x for the missing width.
Debora [2.8K]3 years ago
4 0
U can use  real world numbers and problems to help 
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Solve 152x = 36. Round to the nearest ten-thousandth. 1.7509 2.6466 0.6616 1.9091
Vlad [161]
The answer is 0.6616
Hope it helped :)
6 0
3 years ago
Read 2 more answers
Carrie wished to build a rectangular dog run along theside of her garage. The garage will serve as one side of the fence. If she
Nat2105 [25]

Given:

Carrie has 180 ft of the fencing and wishes the fence to be 4times as long as it is wide.

To Find:

The area in square feet that the fencing encloses.

Answer:

The fence encloses 3600 sq ft.

Step-by-step explanation:

Let x denote the length of the dog run and y denote the width of the dog run.

Given that the garage wall serves as one side of the dog run, we are left with 3 other sides to instal the fence.

Carrie has 180ft of fencing with her, so the sum of the lengths of the 3 sides has to be equal to 180. We can represent this in the form of an equation as

x+y+y=180\\\\x+2y=180\\\\x=180-2y

We are also given that Carrie wishes the fence to be 4 times as long as it is wide.

So,

x=4y

Replacing this value into the first equation, we have

4y=180-2y\\\\
6y=180\\\\
y=30

Therefore,

x=(4)(30)=120

Thus, the length of the dog run is 120ft and the width is 30ft.

The area enclosed will be equal to the length multiplied with the width. So,

(120)(30)=3600

The fence encloses 3600 sq ft.


4 0
3 years ago
Could I have some help on my patterns assignment​
nydimaria [60]

Answer:

1. 13, 18, 23, 28, 33, 38, 43, 48, 53

<em>Conjecture:</em> Each term is 5 more than the previous term.

2. 512, 256, 128, 64, 32, 16, 8, 4, 2

<em>Conjecture:</em> Each term is half of the previous term.

3. 1, 8, 27, 64, 125, 216, 343, 512, 729

<em>Conjecture:</em> Each term is a cube. To find the nth term, cube n.

4. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

<em>Conjecture:</em> Each term is the next consecutive prime number.

5. <u><em>Correct as is</em></u>

6. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89

<em>Conjecture:</em> Each term is the sum of the 2 previous numbers.

Step-by-step explanation:

1.

Given: 13, 18, 23, 28

<em>We add 5 to each number</em>

13 + 5 = 18

18 + 5 = 23

23 + 5 = 28

28 + 5 = 33

33 + 5 = 38

38 + 5 = 43

43 + 5 = 48

48 + 5 = 53

2.

Given : 512, 256, 128, 64

<em>Divide each number by 2</em>

512 ÷ 2 =256

256 ÷ 2 = 128

128 ÷ 2 = 64

64 ÷ 2 = 32

32 ÷ 2 = 16

16 ÷ 2 = 8

8 ÷ 2 = 4

4 ÷ 2 = 2

3.

Given: 1, 8, 27, 64

<em>We have to cube n:</em>

<u><em>n³ = consecutive number cubed</em></u>

<u><em></em></u>

1³ = 1 or 1 × 1 × 1

2³ = 8 or 2 × 2 × 2

3³ = 27 or 3 × 3 × 3

4³ = 64 or 4 × 4 × 4

5³ = 125 or 5 × 5 × 5

6³ = 216 or 6 × 6 × 6

7³ = 343 or 7 × 7 × 7

8³ = 512 or 8 × 8 × 8

9³ = 729 or 9 × 9 × 9

4.

Consecutive prime number: <u><em>Consecutive prime numbers are those that have no gaps or prime numbers between them.</em></u>

Look at the picture in the link:

6.

Given: 1, 1, 2, 3, 5, 8

<em>We must add the two previous numbers to get the next term:</em>

1 + 1 = 2

1 + 2 = 3

2 + 3 = 5

3 + 5 = 8

5 + 8 = 13

8 + 13 = 21

13 + 21 = 34

21 = 34 = 55

34 + 55 = 89

8 0
1 year ago
What is the interquartile range of this data set 2, 5, 9, 11, 18, 30, 42, 48, 71, 73, 81
Shtirlitz [24]

Answer:

I believe the answer is 62.

Step-by-step explanation:

Hope my answer has helped you!

8 0
3 years ago
The function d(s) = 0.0056s squared + 0.14s models the stopping distance
victus00 [196]

Answer:

The car must have a speed of 25 kilometres per hour to stop after moving 7 metres.

Step-by-step explanation:

Let be d(s) = 0.0056\cdot s^{2} + 0.14\cdot s, where d is the stopping distance measured in metres and s is the speed measured in kilometres per hour. The second-order polynomial is drawn with the help of a graphing tool and whose outcome is presented below as attachment.

The procedure to find the speed related to the given stopping distance is described below:

1) Construct the graph of d(s).

2) Add the function d = 7\,m.

3) The point of intersection between both curves contains the speed related to given stopping distance.

In consequence, the car must have a speed of 25 kilometres per hour to stop after moving 7 metres.

4 0
2 years ago
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