GingerAle

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GA

Solution:

\(\text {Let the number be (n) then } \cdots\\ \small \text { }\\ \underbrace {\small \text {Twice the square of a number}} _ {\large \bf 2n^2 }\ \underbrace {\small \text {increased by}} _ {\large \bf +}\ \underbrace {\small \text {five times the product of the number and three more than the number}} _ {\large \bf 5(n)(n+3)}\\ \\ \underbrace {\small \text {is}} _ {\large \bf =} \underbrace {\small \text {the opposite of}} _ {\large \bf -} \underbrace {\small \text { the quotient of six and the number}} _ {\large \bf 6/n}\\ \text{ }\\ \text{ }\\ \hspace {7cm} \bf 2n^2+5(n)(n+3) = -6/n \\ \)

GA

Solution _{^{(using a mathematical construction)}}:

\(\text {The prime factors of } 3240= 2^3 \cdot 3^4 \cdot 5^1 \\ \text {Assemble the primes as multiples of three (3), and count each construction. } \\ \text {For 3240: }\\ \hspace{.8cm} \text {There can be (0, 1, 2, or 3) twos (2) in the construction of a divisor. }\\ \hspace{1.cm} \small \text { So there are a total of four (4) ways to use a two (2) –one of the ways is to} \textbf { not} \text { use it. } \\ \small \text { }\\ \hspace{.8cm} \text {There can be (1, 2, 3 or 4) threes (3) in the construction of a divisor. }\\ \hspace{1.cm} \small \text {Note that for a number to be a multiple of three (3), the number must have at least one multiple of three (3) - }\\ \hspace{1 cm} \small \text {that is, there cannot be a zero (0) number of threes (3). }\\ \hspace{1.cm} \small \text {So there are a total of four (4) ways to use a three (3). }\\ \text { }\\ \hspace{.8cm} \text {There can be (0 or 1) fives (5) in the construction of a divisor. }\\ \hspace{1.cm} \small \text { So there are a total of two (2) ways to use a five (5) –one of the ways is to not use it.}\\ \text { }\\ \text {The product of these counts is } 4\cdot 4\cdot 2 = 32 \\ \text { }\\ \text {There are } \textbf {32 divisors of 3240 that are multiples of three (3). }\\ \)

GA

Hi Melody,

Solution:

\( \small \text {Translate Joules to Liter-atmospheres}\\ w = -5000J * \dfrac{ 1L * atm} {101.325J} =-49.35L*atm \,| \small \text { Negative (-) means energy entered system.}\\ w = −P*\Delta V\\ \small \text {Change in volume }\\ \Delta V = -\dfrac {w}{P} = -\dfrac {-49.35L*atm }{1 atm}=+ 49.35L \,| \small \text { Positive (+) means gas expanded.}\\ \text { }\,\\ \text {Final Volume of the cylinder } 49.35L\)

GA

Solution:

\(\text {Hugo can receive a maximum of 600 points for all tests. }\\ \text {The lowest “B” score is } (0.80 * 600) = 480 \text { points. }\\ \text {The highest “B” score is } (0.899 * 600) =539 \text { points. }\\ \text {Total points for the first four tests are } (82) + (88) + (91) + (83) = 344 \\ \text {For a “B” grade, the final score (F) must have a range of }\\ (479-344) \ge (F) \lt (540-344)\\ \hspace{2.6em} (135) \ge (F) \lt (196)\\ \; \\ \text {Answer “K” exceeds this range. }\\ \)

GA

Solution:

_{(Adapted from dialogues with Lancelot Link)}

\(\text {Note the “handshake” formula: n(n – 1)/2 = h where (n) is the number of persons and (h) is the number of handshakes. }\\ \text {In this question the number of handshakes is depicted by }\\ \left . y + \dfrac {(xy)(xy-1)}{2} = 281 \right | \text { where (x) = number of teams, (y) = number of gymnasts. }\\ \text {Using (n) as the product of (xy) gives } y + \dfrac {n(n-1)}{2} = 281\\ \text {Find the largest integer for (n), such that }\dfrac {n(n-1)}{2} \lt 281\\ \tiny \text {(If you want to be lazy, look up OEIS A000217 “Triangular” numbers and count them. Or ...)}\\ \text {To “zero” in on the largest integer, choose a few arbitrary values for (n). }\\ n=20; n(n-1)/2 = 190\\ n=25; n(n-1)/2 = 300\\ n=24; n(n-1)/2 = 276 \leftarrow \text { This one. }\\ \;\\ y + 276 = 281\\ y = 281 – 276\\ y = 5\\ \)

GA

That’s so true.

I’m collecting her mathematical works of art. After I have enough, I might publish them.

How do these sound for titles?  

Ladders Out of Limbo: Saint Mary’s Escape Routes from Geometric Purgatory

and

From Abomination to Algebra: Saint Mary’s Progressive Solutions to Parochial Problems

These kinda have a Notre Dame du Lac, math goddess tone to it.

GA

Honestly! Sometimes you are just a big pain in the forum’s “assets!” _{(Also, you have a wrong word: “peace.” Very sloppy!)}

Worst of all, you didn’t send “Happy Birthday” greetings to Hectictar.