F=ir^t
139=134r^10
139/134=r^10
r=(139/134)^(1/10) then:
f=134(139/134)^(t/10) so in 2014, t=24 so
f=134(139/134)^(2.4)
f≈146 million (to nearest million)
Some will say that you have to use the exponential function, but it really gives you the same answer...even for continuous compounding :)...
A=Pe^(kt)
139=134e^(10k)
139/134=e^(10k)
ln(139/134)=10k
k=ln(139/134)/10 so
A=134e^(t*ln(139/134)/10) when t=24
A=134e^(2.4*ln(139/134))
A≈146 million (to nearest million)
The only real reason or advantage to using A=Pe^(kt) is when you start getting into differential equations...
Answer:
Step-by-step explanation:
Answer:
1. 21x⁴+3y-35x² + 41
2. -21x⁴-3y+6x² + x
Step-by-step explanation:
When adding and subtracting polynomials , you can use the distributive property to add or subtract the coefficients of like terms.
1. The polynomial is 21x⁴ + 3y -6x² + 34
To obtain polynomial 29x² -7 , we must subtract some polynomial from it.
Let that polynomial be k.
So, 21x⁴ + 3y -6x² + 34 - k = 29x² -7
k = 21x⁴ + 3y - 6x² +34 - 29x² +7 = 21x⁴ + 3y - 35x² + 41
2. To obtain a first degree polynomial, let that polynomial be x +34
So, 21x⁴ + 3y - 6x² + 34 + K = x + 34
K = x + 34 - 21x⁴ -3y + 6x² - 34
= -21x⁴ - 3y + 6x² + x
Answer:
Subsets of the given set = 32
Proper subset of the given set = 31
Step-by-step explanation:
Given set is {18, 8, 14, 9, 6} having 5 elements.
We know number of subsets of a set having n elements are represented by
where n is the number of elements in the set.
Therefore, subsets of the given set = 
= 32
Since original set itself is a subset of its own, is not a proper subset.
Therefore, proper subset of a set = 
= 
= 32 - 1
= 31
It could be B but I want to say D because it relates more to the question.
