In general 2^n is the number of pieces of information n bits can store.
(a) Antonio can use the formula 2^n-1 to indicate the maximum size of numbers an n-bit word can store.
(b) for a 16-bit system, the maximum number of keys Millie can collect is 2^(16)-1 = 65536-1=65535
(c) For a 14 byte program, there are 14*8=112 bits.
Answer:
The correct options are;
1) Write tan(x + y) as sin(x + y) over cos(x + y)
2) Use the sum identity for sine to rewrite the numerator
3) Use the sum identity for cosine to rewrite the denominator
4) Divide both the numerator and denominator by cos(x)·cos(y)
5) Simplify fractions by dividing out common factors or using the tangent quotient identity
Step-by-step explanation:
Given that the required identity is Tangent (x + y) = (tangent (x) + tangent (y))/(1 - tangent(x) × tangent (y)), we have;
tan(x + y) = sin(x + y)/(cos(x + y))
sin(x + y)/(cos(x + y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y))
(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y) - sin(x)·sin(y)) = (Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y))
(Sin(x)·cos(y) + cos(x)·sin(y))/(cos(x)·cos(y))/(cos(x)·cos(y) - sin(x)·sin(y))/(cos(x)·cos(y)) = (tan(x) + tan(y))(1 - tan(x)·tan(y)
∴ tan(x + y) = (tan(x) + tan(y))(1 - tan(x)·tan(y)
Answer:
45 degrees
Step-by-step explanation:
180 - 60 - 75 = 45
Angle RVE is the same as angle LVA
Answer:
since R seems to be in middle between ST
ST=3+1=4
Step-by-step explanation:
