9514 1404 393
Answer:
5) 729, an=3^n, a[1]=3; a[n]=3·a[n-1]
6) 1792, an=7(4^(n-1)), a[1]=7; a[n]=4·a[n-1]
Step-by-step explanation:
The next term of a geometric sequence is the last term multiplied by the common ratio. (This is the basis of the recursive formula.)
The Explicit Rule is ...
for first term a₁ and common ratio r.
The Recursive Rule is ...
a[1] = a₁
a[n] = r·a[n-1]
__
5. First term is a₁ = 3; common ratio is r = 9/3 = 3.
Next term: 243×3 = 729
Explicit rule: an = 3·3^(n-1) = 3^n
Recursive rule: a[1] = 3; a[n] = 3·a[n-1]
__
6. First term is a₁ = 7; common ratio is r = 28/7 = 4.
Next term: 448×4 = 1792
Explicit rule: an = 7·4^(n-1)
Recursive rule: a[1] = 7; a[n] = 4·a[n-1]
9. D because it fits the description when it is graphed.
11. D because it has the correct vertex, which is (1.5, 4.75).
12. f(x)=-4(x-2)²+16 works with the table and provided graph!
13. 0 isn't a zero for f(x), so B is correct (the zeros are 1/2 and 4).
14. The width is about 3.831.
15. The answer is B because the equation for a parabola in this form is:
f(x) = (x - r₁)(x - r₂)
Every one of these except for 15 was solved using desmos, a graphing calculator.
Hope this helps!
Answer:
Some answers to the reducing fraction sheet
Q14 = 4/5 because divide 12 and 15 by 3
Q15=can't be reduced because 11 is a prime number so it can't be divided by anything
Q17=can't be reduced because 17 is a prime number
Q18=can't be reduced because 13 is a prime number
Q19=38/9 because 114 and 27 would be divided by 3
Q20= -1/1 because you would divide both 14,529 by itself would equal to - 1/1
I can only help you with this sheet I can't help you with the other one hope this was helpful! :)
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>
