A triangle should have 3 points and 3 sides. Let say that the point is ABC. Then the sides would be AB, AC and BC.
There are 3 strings with a different length that can be put into the sides. Assuming the string can be used once, then the possible way would be:
3!/(1+3-3)!= 3!/1!= 3*2*1= 6 ways
Answer:
Its 4 thats the answer
Step-by-step explanation:
let me know if this helps
Answer:
Step-by-step explanation:
Remark
The editor must have brackets put around the denominator when there are 2 terms.
That means I think the question is (√5) / (√8 - √3). If this is incorrect, leave a note.
To rationalize the denominator, you must multiply numerator and denominator by the conjugate (√8 + √3).
Solution
√5 * (√8 - √3) / ( (√8 - √3) * (√8 + √3) )
I don't think there is any point in removing the brackets in the numerator. Just leave it.
The denominator is a different matter.
denominator = ( (√8 - √3) * (√8 + √3) )
√8 * √8 = 8
√8 * √3 = √24
- √3 * √8 = - √24
-√3 * √3 = - 3
Take a close look at the 2 middle terms. They cancel out because one of them is plus and the other minus.
What you are left with is 8 - 3 = 5
So the final answer is
√5 * (√8 - √3)
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5
0.04 is multiplied 10 times to get 0.004
15.
if he ordered the same amount each month (x) and it is a whole number,
then
12*x=6150, and
x =6150/12 should be a whole number
x=512.5
so 512.5 is not a whole number and the produce company does not agree
16.
16410/138≈118.91
because it not a whole number, she sells pieces for different prices
