Answer:
a_n=-\frac{1}{4 a_{n-1}
Step-by-step explanation:
The recursive formula for the geometric sequence is given by:
a_n = a_{n-1} \cdot r
where,
r is the common ratio terms
-16, 4, -1, ...
This is a geometric sequence.
Here, and
Since,
ans so on .....
Substitute the given values we have;
⇒
Therefore, the recursive formula for the following geometric sequence is,
A polynomial is a perfect square trinomial if in form
a^2x^2-2ab+b^2
look at each
we can eliminate A and B since we have -b^2 not +b^2
look at the rest
C. 5^2b-3*5*b+3^2
nope
D. 4^2x^2-2*4*3x+3^3
yes
answe is D (factors to (4x-9)^2)
Answer:
Option D is correct.
The equation with roots 3 plus or minus square root 2 is x² - 6x + 7
Step-by-step explanation:
The roots of the unknown equation are
3 ± √2, that is, (3 + √2) and (3 - √2)
The equation can then be reconstructed by writing these roots as the solutions of the quadratic equation
x = (3 + √2) or x = (3 - √2)
The equation is this
[x - (3 + √2)] × [x - (3 - √2)]
(x - 3 - √2) × (x - 3 + √2)
x(x - 3 + √2) - 3(x - 3 + √2) - √2(x - 3 + √2)
= x² - 3x + x√2 - 3x + 9 - 3√2 - x√2 + 3√2 - 2
Collecting like terms
= x² - 3x - 3x + x√2 - x√2 - 3√2 + 3√2 + 9 - 2
= x² - 6x + 7
Hope this Helps!!!
Answer:
3 also could be said as 3/1
Step-by-step explanation:
3.072 seconds, (38.4/12.5)