Answer: huh?? Need more info.
Step-by-step explanation:
The answer is D. Think back to the quadratic formula
You would plug in values of a, b, and c to find the zeros of the equations. It would the same way in a spreadsheet program where the cells contain the a, b, and c values but you can have more values and a different equation.
Let a = 693, b = 567 and c = 441
Now first we will find HCF of 693 and 567 by using Euclid’s division algorithm as under
693 = 567 x 1 + 126
567 = 126 x 4 + 63
126 = 63 x 2 + 0
Hence, HCF of 693 and 567 is 63
Now we will find HCF of third number i.e., 441 with 63 So by Euclid’s division alogorithm for 441 and 63
441 = 63 x 7+0
=> HCF of 441 and 63 is 63.
Hence, HCF of 441, 567 and 693 is 63.
Answer:
4x/y4
Step-by-step explanation:
2(2x/y4)
2 x 2x/ y4
=4x/y4
