Answer:
2/√10 - 8√2
Step-by-step explanation:
Multiply the numerator and denominator by a root to remove the radical from the numerator.
Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).
Let <em>x</em> be the inverse. Then <em>x</em> is such that
23<em>x</em> ≡ 1 (mod 1000)
Use the Euclidean algorithm to solve for <em>x</em> :
1000 = 43×23 + 11
23 = 2×11 + 1
→ 1 ≡ 23 - 2×11 (mod 1000)
→ 1 ≡ 23 - 2×(1000 - 43×23) (mod 1000)
→ 1 ≡ 23 - 2×1000 + 86×23 (mod 1000)
→ 1 ≡ 87×23 - 2×1000 ≡ 87×23 (mod 1000)
→ 23⁻¹ ≡ 87 (mod 1000)
Answer:
1500
Step-by-step explanation:
10 * 10 = 100
35*40 = 1400
Add them
I have no clue actually this is weird
Answer:
2:12
Step-by-step explanation:
6 and 4 = 10
5 and 5 = 10
