Is it possible to form a triangle with the given side lengths? 15 yd,16yd,30yd
2 answers:
<u>Answer:</u>
<h2>
Yes</h2>
<u>Explanation:</u>
15 < 16+30
16 < 15+30
30 < 16+15
Yes it is possible to form a trial for
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Answer:
1/6 times the number cube wil land on 2
Step-by-step explanation:
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the answer would be d, very much appreciated if you would give me brainliest
Answer: 
<u>Step-by-step explanation:</u>
Isolate w by performing the following steps
- Multiply by 6 on both sides to clear the denominator
- Subtract 3 from both sides
- Divide both sides by 2
![y=\dfrac{1}{2}+\dfrac{w}{3}\\\\\\6\bigg[y=\dfrac{1}{2}+\dfrac{w}{3}\bigg]\quad \implies \quad 6y=3+2w\\\\\\6y-3=3-3+2w\quad \implies \quad 6y-3=2w\\\\\\\dfrac{6y-3}{2}=\dfrac{2w}{2}\quad \implies \quad \large\boxed{\dfrac{6y-3}{2}=w}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7Bw%7D%7B3%7D%5C%5C%5C%5C%5C%5C6%5Cbigg%5By%3D%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7Bw%7D%7B3%7D%5Cbigg%5D%5Cquad%20%5Cimplies%20%5Cquad%206y%3D3%2B2w%5C%5C%5C%5C%5C%5C6y-3%3D3-3%2B2w%5Cquad%20%5Cimplies%20%5Cquad%206y-3%3D2w%5C%5C%5C%5C%5C%5C%5Cdfrac%7B6y-3%7D%7B2%7D%3D%5Cdfrac%7B2w%7D%7B2%7D%5Cquad%20%5Cimplies%20%5Cquad%20%5Clarge%5Cboxed%7B%5Cdfrac%7B6y-3%7D%7B2%7D%3Dw%7D)
If you just divide both you get 9.7
