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
I believe this is scale factor? SF=new/old. 10/8= a scale factor of 1.25 (the image gets 1.25x bigger). To find x, multiply 5 by 1.25 and subtract the original length of 5 (you subtract this length because x is only the value of the new part of the triangle). X=6.25-5, X=1.25. Try to find y yourself using this scale factor. (Brainliest would be appreciated)
Lets round it to the nearest ten
A 97 ====> 100
B 118 ===> 120
C 179 ===> 180
D 5091 ==> 5090
No result yet, lets round to the nearest hindred.
A 97 ====> 100
B 118 ===> 100
C 179 ===> 180
D 5091 ==> 5100
As we can see only A give the same result when we round it to the nearest hundred and nearest ten.
Combine like terms
-2x-4=-2x-4
Since the equations are the same, the answer is infinitely many solutions.
I’m pretty sure you add 78+55= 133
180-133=47
so 9x+2=47
subtract 2 from 47 and itself
so 9x=45
divide 9 and 45 by 9
so x=5
