Given that:
James is 6 times older than his friend Zach, if Zach added 8 to his age and multiplied this sum by 3 .
To find:
The equation that can be used to solve for zach's age.
Solution:
Let the age of zach be Z.
Zach added 8 to his age = Z+8
And multiplied this sum by 3 = 3(Z+8)
James is 6 times older than his friend Zach = 6Z
Now,
Divide both sides by 3.
Therefore, the required equation is is age of zach is 8 years.
A(n) = -5 + 6(n - 1)a(n)=−5+6(n−1)a, left parenthesis, n, right parenthesis, equals, minus, 5, plus, 6, left parenthesis, n, min
DENIUS [597]
Answer:
The 12th term is 61
Step-by-step explanation:
I will assume that your a(n) = -5 + 6(n - 1) is correct; the rest is redundant (duplicative, unneeded).
To find the 12th term, substitute 12 for n in the above formula:
a(12) = -5 + 6(12 - 1) = -5 + 6(11) = 66 - 5, or 61
The 12th term is 61
Answer:
a = - 1, b = - 3, c = - 5, d = - 6
Step-by-step explanation:
Substitute the appropriate values of x into the equation and evaluate
x = - 3 : y = - (- 3) - 4 = 3 - 4 = - 1 → a
x = - 1 : y = - (- 1) - 4 = 1 - 4 = - 3 → b
x = 1 : y = - 1 - 4 = - 5 → c
x = 2 : y = - 2 - 4 = - 6 → d
Work:
624 divided by 12 equals 52
answer:
52 dozens
Proving a relation for all natural numbers involves proving it for n = 1 and showing that it holds for n + 1 if it is assumed that it is true for any n.
The relation 2+4+6+...+2n = n^2+n has to be proved.
If n = 1, the right hand side is equal to 2*1 = 2 and the left hand side is equal to 1^1 + 1 = 1 + 1 = 2
Assume that the relation holds for any value of n.
2 + 4 + 6 + ... + 2n + 2(n+1) = n^2 + n + 2(n + 1)
= n^2 + n + 2n + 2
= n^2 + 2n + 1 + n + 1
= (n + 1)^2 + (n + 1)
This shows that the given relation is true for n = 1 and if it is assumed to be true for n it is also true for n + 1.
<span>By mathematical induction the relation is true for any value of n.</span>
