A person places $96300 in an investment account earning an annual rate of 2.8%, compounded continuously. Using the formula V=PertV = Pe^{rt} V=Pe rt , where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 7 years. Substituting our given numbers in where P = 96,300, r = 0.028, and t = 7, we get: V = 96,300 * e^(0.028 * 7) V = 96,300 * e^0.196 V = 96,300 * 1.21652690533 V = [B]$117,151.54[/B]

list the natural numbers less than 70 that are divisible by 8 Natural numbers are {1, 2, 3, ... We want natural numbers less than 70 which are divisible by 8: [LIST] [*]8 * 1 = 8 [*]8 * 2 = 16 [*]8 * 3 = 24 [*]8 * 4 = 32 [*]8 * 5 = 40 [*]8 * 6 = 48 [*]8 * 7 = 56 [*]8 * 8 = 64 [/LIST] Our answer is: [B]{8, 16, 24, 32, 40, 48, 56, 64}[/B]

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natural numbers that are factors of 16 Natural numbers are positive integers starting at 1. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Of these, [URL='https://www.mathcelebrity.com/factoriz.php?num=16&pl=Show+Factorization']the only factors of 16[/URL] are: {[B]1, 2, 4, 8, 16}[/B]

Olga wrote all the natural numbers from 1 to k. Including 1 and k. How many numbers did she write? The formula for the number of numbers including A to B is: B - A + 1 With A = 1 and B = k, we have: k - 1 + 1 [B]k[/B]

P is the natural numbers that are factors of 25 we type in [I][URL='https://www.mathcelebrity.com/factoriz.php?num=25&pl=Show+Factorization']factor 25[/URL][/I] into our math engine and we get: {1, 5, 25} Since [U]all[/U] of these are natural numbers, our answer is: [B]{1, 5, 25}[/B]

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Suppose x is a natural number. When you divide x by 7 you get a quotient of q and a remainder of 6. When you divide x by 11 you get the same quotient but a remainder of 2. Find x. [U]Use the quotient remainder theorem[/U] A = B * Q + R where 0 ? R < B where R is the remainder when you divide A by B Plugging in our numbers for Equation 1 we have: [LIST] [*]A = x [*]B = 7 [*]Q = q [*]R = 6 [*]x = 7 * q + 6 [/LIST] Plugging in our numbers for Equation 2 we have: [LIST] [*]A = x [*]B = 11 [*]Q = q [*]R = 2 [*]x = 11 * q + 2 [/LIST] Set both x values equal to each other: 7q + 6 = 11q + 2 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=7q%2B6%3D11q%2B2&pl=Solve']equation calculator[/URL], we get: q = 1 Plug q = 1 into the first quotient remainder theorem equation, and we get: x = 7(1) + 6 x = 7 + 6 [B]x = 13[/B] Plug q = 1 into the second quotient remainder theorem equation, and we get: x = 11(1) + 2 x = 11 + 2 [B]x = 13[/B]

the set of natural numbers less than 7 that are divisible by 3 Natural Numbers less than 7 {1, 2, 3, 4, 5, 6} Only 2 of them are divisible by 3. Divisible means the number is divided evenly, with no remainder: [B]{3, 6}[/B]

the sum of 3 consecutive natural numbers, the first of which is n Natural numbers are counting numbers, so we the following expression: n + (n + 1) + (n + 2) Combine n terms and constants: (n + n + n) + (1 + 2) [B]3n + 3 Also expressed as 3(n + 1)[/B]

the sum of 3 consecutive natural numbers, the first of which is n We have: n + (n + 1) + (n + 2) Grouping like terms, we have: [B]3n + 3[/B]

The sum of 3 consecutive natural numbers, the first of which is n. We have 3 numbers: n, n + 1, and n + 2 Add them up: n + (n + 1) + (n + 2) Group like terms: [B]3n + 3[/B]

I saw this ticket come through today. The answer is x > 6. Natural numbers are positive numbers not 0. So 1, 2, 3, ... Let me build this shortcut into the calculator. Also, here is the[URL='http://www.mathcelebrity.com/interval-notation-calculator.php?num=x%3E6&pl=Show+Interval+Notation'] interval notation[/URL] for that expression.