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.
Just divide $100 by 5. 100 divided by 5 = $20 each.
Answer:
x₁ = -4
x₂ = 3
Step-by-step explanation:
x²+ x + 12 = 0
x = {-1±√((1²)-(4*1*-12))} / (2*1)
x = {-1±√(1+48)} / 2
x = {-1±√49} / 2
x = {-1±7} / 2
x₁ = {-1-7} / 2 = -8/2 = -4
x₂ = {-1+7} / 2 = 6/2 = 3
Check:
x₁
-4² + (-4) - 12 = 0
16 - 4 - 12 = 0
x₂
3² + 3 - 12 = 0
9 + 3 - 12 = 0
Answer:
p= 1/3 or 0.33 repeating
Step-by-step explanation:
5-19p=-p-1
5=18p-1
6=18p
divide by 18
6/18 = 1/3 or 0.33 repeating
