The correct answer is D) Miguel: d=23t+10, Gabby d=28t
We multiply the speed of each person by t, the amount of time. This is where the 23t in Miguel's equation and the 28t in Gabby's equation come from. The starting point is added to this, which leads to 23t+10 and 28t+0 or 28t.
Hey there!
(6^3 * 2^6) / 2^3
= (6 * 6 * 6 * 2 * 2 * 2 * 2 * 2 * 2) / 2 * 2 * 2
= (36 * 6 * 4 * 4 * 4) / 4 * 2
= (216 * 16 * 4) / 8
= 3,456 * 4 / 8
= 13,824 / 8
= 1,728
Looking for something that gives you the result of: 1,728
Option A.
12^3
= 12 * 12 * 12
= 144 * 12
= 1,728
Option A. is. possible answer
Option B.
6^3
= 6 * 6 * 6
= 36 * 6
= 216
216 ≠ 1,728
Option B. is incorrect
Option C.
12^6
= 12 * 12 * 12 * 12 * 12 * 12
= 144 * 144 * 144
= 20,736 * 144
= 2,985,984
2,985,984 ≠ 1,728
Option C. is also incorrect
Option D.
2^6 * 2^3
= 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2
= 4 * 4 * 4 * 4 * 2
= 16 * 16 * 2
= 256 * 2
= 512
512 ≠ 1,728
Option D. is also incorrect
Option E.
2^3 * 3^3
= 2 * 2 * 2 * 3 * 3 * 3
= 4 * 2 * 9 * 3
= 8 * 27
= 216
216 ≠ 1,728
Option E. is also incorrect.
Therefore, the answer should be:
Option A. 12^3
Good luck on your assignment & enjoy your day!
~Amphitrite1040:)
A x B x C = 0
Where A and B are integers > 10
Any value multiplied by 0 = 0
Therefore C = 0
Check the picture below.
well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.
bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.
![\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}](https://tex.z-dn.net/?f=%5Cbf%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B1%7D-%5Cstackrel%7By1%7D%7B4%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B5%7D-%5Cunderset%7Bx_1%7D%7B1%7D%7D%7D%5Cimplies%20%5Ccfrac%7B-3%7D%7B4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bslope%20of%20AB%7D%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7B%5Cunderline%7Bnegative%20reciprocal%7D%20and%20slope%20of%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%7D%7B3%7D%7D)
so, it passes through the midpoint of AB,
so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)
