Answer: C
<u>Step-by-step explanation:</u>
∑ 8(
)⁽ⁿ⁻¹⁾ from n = 1 to n = 5
n = 1 → 8()⁽¹⁻¹⁾
= 8 = 
n = 2 → 8()⁽²⁻¹⁾
= 
= 
n = 3 → 8()⁽³⁻¹⁾
= 
= 
n = 4 → 8()⁽⁴⁻¹⁾
= 
= 
n = 5 → 8()⁽⁵⁻¹⁾
= 
=
Sum = 
= 408.99
Answer:
rate
Step-by-step explanation:
Let x be the 1st odd number, and x+2 the second odd consecutive number:
(x)(x + 2) = 6[((x) + (x+2)] -1
x² + 2x = 6(2x + 2) - 1
x² + 2x = 12x +12 - 1
And x² - 10x - 11=0
Solve this quadratic expression:
x' = [+10 +√(10²- 4.(1)(-11)]/2 and x" = [+10 -√(10²- 4.(1)(-11)]/2
x' = [10 + √144]/2 and x" = [10 - √64]/2
x' = (10+12)/2 and x" = (10-12)/2
x = 11 and x = -1
We have 2 solutions that satisfy the problem:
1st for x = 11, the numbers at 11 and 13
2nd for x = - 1 , the numbers are -1 and +1
If you plug each one in the original equation :(x)(x + 2) = 6[((x) + (x+2)] -1
you will find that both generates an equlity
Answer:
Step-by-step explanation:
Unclear. Did you mean f(x) = 2^(x + 3)? If so, the parentheses are mandatory.
Assuming that f(x) = 2^(x + 3) is correct, then:
a) f(2) = 2^(2 + 3) = 2^5 = 32
b) f'(x) = 2^(x + 3)*(x + 3)' = 2^(x + 3)(1) = 2^(x + 3)
c) f"(x) = 2^(x + 3) (same as in (b)), so that:
f"(12) = 2^15
Answer:40.00
Step-by-step explanation:just add
