Mp. 146.54
dis . 15%
sp= 146.54 - 15 % of 146.54
=124.559
Ok, so the formula for compound QUARTERLY is A=P(1+ʳ/n)ⁿ<span>ᵗ.
P= The initial amount.
R= The Rate
T= The Time/Number of Years
N= Number of time interest is compounded per year. (In this case its 4 because it compounded QUARTERLY.)
So if you input the numbers, you will get A=10,000(1+0.0625/4)</span>⁴⁽²⁵⁾<span>
Now solve inside the parenthesis.
10,0 00(1.0625/4)</span>⁴⁽²⁵⁾
Now you will need a calculator for the next part...
Do 1.0625/4 and times it by 10,000 .
You will get 0.0265265. You can't for get about the ⁴⁽²⁵⁾.
⁴⁽²⁵⁾=<span>¹⁰⁰
</span>2,656.25¹⁰⁰=
$47134.43 Hope This Helped!
Answer:
Lets say that P(n) is true if n is a prime or a product of prime numbers. We want to show that P(n) is true for all n > 1.
The base case is n=2. P(2) is true because 2 is prime.
Now lets use the inductive hypothesis. Lets take a number n > 2, and we will assume that P(k) is true for any integer k such that 1 < k < n. We want to show that P(n) is true. We may assume that n is not prime, otherwise, P(n) would be trivially true. Since n is not prime, there exist positive integers a,b greater than 1 such that a*b = n. Note that 1 < a < n and 1 < b < n, thus P(a) and P(b) are true. Therefore there exists primes p1, ...., pj and pj+1, ..., pl such that
p1*p2*...*pj = a
pj+1*pj+2*...*pl = b
As a result
n = a*b = (p1*......*pj)*(pj+1*....*pl) = p1*....*pj*....pl
Since we could write n as a product of primes, then P(n) is also true. For strong induction, we conclude than P(n) is true for all integers greater than 1.
Answer:
I can not answer this question because vital information is missing.
Step-by-step explanation:
Pleas add a picture. At this moment I can not answer this question because vital information is missing.
Answer:
(Please vote me Brainliest if this helped!)
Step-by-step explanation:
1. 
2. 
3. 
4. 
5. 
