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What is the value of n if 2n plus 7 equals 19?
2n plus 7 equals 19 2n equals 12 n equals 6
What is 2n - 7 equals -1?
2n - 7 = - 1 add 7 to each side 2n - 7 + 7 = - 1 + 7 2n = 6 divide each sides integer by 2 (2/2)n = 6/2 n = 3 ---------------check in original equation 2(3) - 7 = - 1 6 - 7 = - 1 - 1 = - 1 ----------------checks and equation is satisfied
What is the nth term of this sequence 3 5 7 9 11?
The nth term of the sequence is 2n + 1.
What is 2n 3n 7-41?
2n+3n+7=-41
Identify the first step to solve the given equation 2n 3 7?
5n-3
What is the value of n if 2n plus 7 equals 19?
2n plus 7 equals 19 2n equals 12 n equals 6
What is 3n plus 3 equals n plus 7?
3n+3=n+7 3n+3-n=n+7-n (subtract n from both sides) 2n+3=7 2n+3-3=7-3 (subtract 3 from both sides ) 2n=4 2n/2=4/2 (divided by 2 on both the sides) n=2 Answer: n=2
What is the Nth term of 2n plus 3?
The Nth term of tn = 2n + 3 is 2N + 3. Replace (substitute) the n by the term number to get its value. t1 = 2 x 1 + 3 = 5 t2 = 2 x 2 + 3 = 7 t3 = 2 x 3 + 3 = 9 etc.
What is 7-2n equals n-14?
7-2n=n-147-2n+2n=n-14+2n7=3n-147+14=3n21=3n21/3=3n/37=n
What is the nth term of 5 7 9?
2n+3. If 5 is the first term, then it is 2n + 3 (2×1 = 2 + 3 = 5 and 2×2 + 3 = 7)
What is 2n - 7 equals -1?
2n - 7 = - 1 add 7 to each side 2n - 7 + 7 = - 1 + 7 2n = 6 divide each sides integer by 2 (2/2)n = 6/2 n = 3 ---------------check in original equation 2(3) - 7 = - 1 6 - 7 = - 1 - 1 = - 1 ----------------checks and equation is satisfied
What is an equation for this problem Three less than twice of what number is -7?
2n-3=(-7) 2n=-4 n=-2
2n plus 19 equals -7?
2n + 19 = -7 2n = -7 - 19 2n = -26 n= -13
What are 4 positive odd numbers whose sum of their squares equals 164?
Answer is 3, 5, 7, 9detailsassume the numbers are2n-3, 2n-1, 2n+1, 2n+3 ......................... (1)(2n-3) 2 + (2n-1) 2 + (2n+1) 2 + (2n+3) 2 = 16416n 2 + 20 = 16416n 2 = 144n2 = 9n = 3Substitute in eq 1 we get the answer above
Why are 3 5 and 7 are the only three consecutive odd prime numbers?
This is because at least one of them must be divisible by 3. The outline proof is as follows: Consider the consecutive odd numbers, 2n+1, 2n+3, 2n+5 ... Suppose 2n+1 is divisible by 3. That is 2n+1 = 3k for some integer k, then 2n+7 = (2n+1)+6 = 3k+6 = 3*(k+2) so that 2n+7 is also divisible by 3. So the run stops with 2n+3, 2n+5. Suppose, instead, that 2n+1 leaves a remainder of 1 when divided by 3. Then 2n+3 and 2n+9 etc are divisible by 3. Again leaving a run of only two consecutive odd primes. Finally, suppose that 2n+1 leaves a remainder of 2 when divided by 3. Then 2n+5, 2n+11 etc are divisible by 3 leaving runs of two consecutive odd primes.
What is 5 over 2n plus 3 over 3n?
15/6n + 6/6n = 21/6n = 7/2n
What is the nth term of this sequence 3 5 7 9 11?
The nth term of the sequence is 2n + 1.
