Answer:
7 Weeks
Step-by-step explanation:
If she starts off with 40 reps for the first week, and she adds 6 reps a week, until she gets to 82 reps a week, then it would be<em> 40 + 6 x answer = 82</em>. Now, there are a couple ways you can do this, but the easiest but not shortest way is to do it one by one.
so, <em>40+6x1= 46</em>
<em>40+6x2= 52</em>
<em>40+6x3= 58</em>, and so on and so forth, until you get to <em>40+6x7</em><u>,</u> which equals 82, therefore giving you the answer of 7, but the way quicker way is to take the total (82) and subtract the number of reps she starts out with (40), then divide that number by the amount that it increases each week (6), so:
(82-40)÷6, which will also give you the answer of 7, or 7 weeks.
I like to start by dividing the given number by 2, unless a larger factor is evident.
1386 = 2(693). It's pretty obvious that we can divide 693 by 3: 693=3(231).
Then we have 1386 = (2)(3)(231). Try the same procedure. Does 231 have any integer, prime factors? See how far you can take this factoring.
Answer:
$8950.37
Step-by-step explanation:
Use the compound amount formula A = P(1 + r/n)^(nt), in which P is the initial amount of money (the principal), r is the interest rate as a decimal fraction, n is the number of times per year that interest is compounded, and t is the number of years.
Here we have A = $11,000, n = 2, r = 0.07 and t = 3, and so:
$11,000 = P(1 + 0.07/2)^(2*3), or
$11,000 = P (1.035)^6
$11,000 $11,000
Solving for P, we get P = ---------------- = ------------- = $8950.37
1.035^6 1.229
Depositing $8950.37 with terms as follows will result in an accumulation of $11,000 after 3 years.
Let the amount deposited (principal) be x, then the amount after the required time = 2x.
A = P(1 + r/n)^nt: where A is the future value = 2x, P is the principal = x, r is the rate = 0.75%, n is the number of accumulation in a year = 12, t is the number of years.
2x = x(1 + 0.0075/12)^12t
2 = (1 + 0.000625)^12t
log 2 = 12t log (1.000625)
log 2 / log (1.000625) = 12t
1109.38 = 12t
t = 92 years
√25