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How do you cancel out exponents?
To cancel out exponents, you can use the property of exponents that states if you have the same base, you can subtract the exponents. For example, in the expression (a^m \div a^n), you can simplify it to (a^{m-n}). Additionally, if you have an exponent raised to another exponent, such as ((a^m)^n), you can multiply the exponents to simplify it to (a^{m \cdot n}). If you set an expression equal to 1, you can also solve for the exponent directly by taking logarithms.
How to simplify n plus 1 factorial divided by n factorial?
Since n! is the product of all the numbers from 1 through n and (n+1)! is everything in n! multiplied by n+1, the quotient is n+1 ■
What does m equals in mn equals m plus n?
m = n/(n-1)
n+m+43+38=180 what is n and m?
N is 98 and M is 1
What is the number of pairs of non-zero integers whose sum is 72 more than their product is?
Let the integers be m and n.∴ By the condition given in the question,m + n - mn = 72∴ m + n - mn = 71 + 1∴ m + n - mn - 1 = 71∴ (m - 1)(1 - n) = 71∴ (m - 1)(1 - n) = 1 × 71 = 71 × 1 = (-1) × (-71) = (-71) × (-1)Case 1:m - 1 = 1 and 1 - n = 71∴ (m, n) = (2, -70)Case 2:m - 1 = 71 and 1 - n = 1∴ (m , n) = (72, 0)This case is inadmissible as m, n are non zero integers.Case 3:m - 1 = -1 and 1 - n = -71∴ (m , n) = (0, 72)This case is inadmissible as m, n are non zero integers.Case 4:m - 1 = -71 and 1 - n = -1∴ (m, n) = (-70, 2)But as the given expression is symmetric, therefore, (2, -70) and (-70, 2) cannot be considered two different pairs.∴ We get only one solution.
Simplify 5m-13n-n plus 4m?
5m-13n-n+4m= (5+4)m+(-13-1)n= 9m-14n
How do you simplify the expression m-2 plus 1-2m plus 1?
m-2+1-2m+1 When simplified: -m
Why is a negative divided by a negative a positive?
if we have two negative numbers, -n and -m and we divide them: (-n)/(-m) then we can rewrite the quotient as : (-1)*n/(-1)*m and... (-1)/(-1) * n/m we know that any number divided by itself is 1 by some rule: so = (-n)/(-m) = n/m which is positive
How to simplify n plus 1 factorial divided by n factorial?
Since n! is the product of all the numbers from 1 through n and (n+1)! is everything in n! multiplied by n+1, the quotient is n+1 ■
What does m equals in mn equals m plus n?
m = n/(n-1)
n+m+43+38=180 what is n and m?
N is 98 and M is 1
What is the opposite of m-n?
The additiove opposite is -m+n or n-m. There is also a multiplicative opposite, which is 1/(m-n)
How do you find equation for a pyramid of oranges stacked 1 2x2 3x3 4x4?
The equation for n layers is S(n) = n(n+1)(2n+1)/6It is simplest to prove it by induction.When n = 1,S(1) = 1*(1+1)(2*1+1)/6 = 1*2*3/6 = 1.Thus the formula is true for n = 1.Suppose it is true for n = m. That is, for a pyramid of m levels,S(m) = m*(m+1)*(2m+1)/6Then the (m+1)th level has (m+1)*(m+1) oranges and soS(m+1) = S(m) + (m+1)*(m+1)= m*(m+1)*(2m+1)/6 + (m+1)*(m+1)= (m+1)/6*[m*(2m+1) + 6(m+1)]= (m+1)/6*[2m^2 + m + 6m + 6]= (m+1)/6*[2m^2 + 7m + 6]= (m+1)/6*(m+2)*(2m+3)= (m+1)*(m+2)*(2m+3)/6= [(m+1)]*[(m+1)+1)]*[2*(m+1)+1]/6Thus, if the formula is true for n = m, then it is true for n = m+1.Therefore, since it is true for n =1 it is true for all positive integers.
What is the number of pairs of non-zero integers whose sum is 72 more than their product is?
Let the integers be m and n.∴ By the condition given in the question,m + n - mn = 72∴ m + n - mn = 71 + 1∴ m + n - mn - 1 = 71∴ (m - 1)(1 - n) = 71∴ (m - 1)(1 - n) = 1 × 71 = 71 × 1 = (-1) × (-71) = (-71) × (-1)Case 1:m - 1 = 1 and 1 - n = 71∴ (m, n) = (2, -70)Case 2:m - 1 = 71 and 1 - n = 1∴ (m , n) = (72, 0)This case is inadmissible as m, n are non zero integers.Case 3:m - 1 = -1 and 1 - n = -71∴ (m , n) = (0, 72)This case is inadmissible as m, n are non zero integers.Case 4:m - 1 = -71 and 1 - n = -1∴ (m, n) = (-70, 2)But as the given expression is symmetric, therefore, (2, -70) and (-70, 2) cannot be considered two different pairs.∴ We get only one solution.
What are the law of exponent?
Exponents are subject to many laws, just like other mathematical properties. These are X^1 = X, X^0 = 1, X^-1 = 1/X, X^m * X^n = X^m+n, X^m/X^n = X^m-n, (X^m)^n = X^(m*n), (XY)^n = X^n * Y^n, (X/Y)^n = X^n/Y^n, and X^-n = 1/X^n.
Can you prove mathematically that a geometric random variable possesses the so called Memory-less property?
Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.
What are 3 consecutive numbers?
They are numbers of the form m, m+1 and m+2 where m is an integer. However, sometimes it can be easier - particularly with an odd number of consecutive integers - to write them as n-1, n and n+1 where n is an integer (= m+1).
