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# The sum of 3 times m and n?

Updated: 4/28/2022

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12y ago

the sum of 3 times m and n

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12y ago
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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The Sum of M and 6

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sum = 2n + 3m

### What is four times the sum of a number and 3?

4(N-3) is the formula, do the math from there and you get "N-3"

10n = -30 n = -3

### What is the average of all the integer's of 13 to 37?

To find the (mean) average, add all the numbers and divide by the number of numbers. The sum of a series of digits (in arithmetic progression, like 13, 14, 15, ... 37) is sum = (first + last) x number_of_digits / 2 So their average is: average = ((first + last) x number_of_digits / 2) / number_of_digits = (first + last) / 2 = average of first and last digits! So the average of the numbers 13, 14, 15, ..., 37 is: average = (13 + 37) / 2 = 50 / 2 = 25 To find the sum of n digits starting with m: Sum = m + (m+1) + ... + (m+n-2) + (m+n-1) Rewrite the sum in reverse order: Sum2 = Sum = (m+n-1) + (m+n-2) + ... + (m+1) + m Add the two sums, term by term: Sum + Sum2 = 2 Sum = (m + (m+n-1)) + (m + (m+n-1)) + ... + (m + (m+n-1)) + (m + (m+n-1)) There are n terms, all (m + (m+n-1)), so: 2 Sum = (m + m+n-1) x n Sum = (m + m+n-1) x n / 2 But n is the number of digits, m is the first number and (m+n-1) is the last, so: Sum = (first + last) x number_of_digits / 2

3 (n + 7) = 16

7(n+3)

### What is the algebraic expression for the sum of 12 and 3 times a number?

Where n = any number, n(12+3)

### How do you Work out average speeds for drag cars?

You record the various speed on the same distance and sum it. Then divide that sum with how many times you record it. Like for x meters : 1. a m/s 2. b m/s 3. c m/s . . n. y m/s The average would be : (a+b+c +...y)/n

### M times N equals 6300 if m and n are 2 digit multiples of 10 what could m and n be?

m and n are 70 and 90

### What is the expresstion of 4 times the difference of a number an 1 is equal to 6 times the sum of a number times 3?

4*(n + 1) = 6*n*3

### The 'nth term of an Arithmetic Progression is 3n-2.Find the sum of first n terms.What is the sum of first 10 terms?

The sum of the 1st n terms is : N(3N-1)/2 Explanation : The sum from 1 to N of (3m-2) = 3 * sumFrom1toN(m) - sumFrom1toN(2) = 3 * (N*(N+1)/2) -2*N = N(3N-1)/2 For N=10 =&gt; 145

m -5 n-3