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boyakko [2]
3 years ago
12

You have 16 yellow beads, 20 red beads, and 24 orange beads to make identical bracelets. What is the greatest number of the brac

elets that you can make using all of these beads?
If answer is correct will mark as brainlest only if it correct tho
Mathematics
1 answer:
zzz [600]3 years ago
6 0

Answer:

4 bracelets

Step-by-step explanation:

We have been given that you have 16 yellow beads 20 red beads and 24 orange beads to make identical bracelet. We are asked to find the greatest numbers of bracelets you can make using all the beads.

To solve our given problem, we need to find greatest common factor of 16, 20 and 24.

Factors of 16: 1, 2, 4, 8, 16.

Factors of 20: 1, 2, 4, 5, 10, 20.

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.

We can see that greatest common factor of 16, 20 and 24 is 4, therefore, you can make 4 bracelets using all the beads.

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PIT_PIT [208]

<u><em>Answer:</em></u>

1. 15.49 cm

2. 14.39 cm

3. 17.32 cm

4. 11.53 cm

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<u><em>Step-by-step explanation:</em></u>

In the given problem, you can solve the unknown side of the triangle using either<em> Pythagorean Theorem</em> or<em> SOA CAH TOA</em> method.<em> </em><em>In Trigo, you can use the Pythagorean Theorem if the triangle has 90 degrees, or it is a right triangle.</em>

In the given problem, the angle of the triangles are unknown, therefore to solve the unknown side of triangles, we will you use the SOA CAH TOA.

<u><em>In this case, since the given side are b and c, we can use CAH (Cos Teta equals Adjacent over Hypotenuse), and TOA ( Tan Teta equals Opposite over Adjacent)</em></u>

Lets provide the solution for the number, and the rest have the same solutions and process.

b= 17, c= 3, let u is the angle

Cosu= 17/23, to get the angle, use inverse cosine.

The result will be 42.34°, and to get the unknown side, use TOA

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It will give you 15.49 cm

In the number 2 and 5, it will be the same when you rotate the triangle.

Read more about the concept of SOA CAH TOA at

<em>brainly.com/question/22767551</em>

<em />

<u><em>#BrainlyFast</em></u>

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