A car is purchased for $24,000 . Each year it loses 30% of its value. After how many years will the car be worth $7300 or less? (Use the calculator provided if necessary.) Write the smallest possible whole number answer. Set up the depreciation equation D(t) where t is the number of years in the life of the car: D(t) = 24,000 * (1 - 0.3)^t D(t) = 24000 * (0.7)^t The problem asks for D(t)<=7300 24000 * (0.7)^t = 7300 Divide each side by 24000 (0.7)^t = 7300/24000 (0.7)^t= 0.30416666666 Take the natural log of both sides: LN(0.7^t) = -1.190179482215518 Using the natural log identities, we have: t * LN(0.7) = -1.190179482215518 t * -0.35667494393873245= -1.190179482215518 Divide each side by -0.35667494393873245 t = 3.33687437943 [B]Rounding this up, we have t = 4[/B]

A natural number greater than 1 has only itself and 1 as factors is called a [B]prime number.[/B]

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]

A super deadly strain of bacteria is causing the zombie population to double every day. Currently, there are 25 zombies. After how many days will there be over 25,000 zombies? We set up our exponential function where n is the number of days after today: Z(n) = 25 * 2^n We want to know n where Z(n) = 25,000. 25 * 2^n = 25,000 Divide each side of the equation by 25, to isolate 2^n: 25 * 2^n / 25 = 25,000 / 25 The 25's cancel on the left side, so we have: 2^n = 1,000 Take the natural log of each side to isolate n: Ln(2^n) = Ln(1000) There exists a logarithmic identity which states: Ln(a^n) = n * Ln(a). In this case, a = 2, so we have: n * Ln(2) = Ln(1,000) 0.69315n = 6.9077 [URL='https://www.mathcelebrity.com/1unk.php?num=0.69315n%3D6.9077&pl=Solve']Type this equation into our search engine[/URL], we get: [B]n = 9.9657 days ~ 10 days[/B]

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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.