The answers is 24, 25, and 26.
A^2+ b^2= c^2
24^2+b^2=25^2
576+b^2=625
Sub 576
b^2= 49 b=7
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)

Answer:
The answer is below
Step-by-step explanation:
The question is not complete. The complete question is:
The 12 foot long bed of a dump truck loaded with debris must rise an angle of 30 degrees before the debris will spill out. Approximately how high must the front of the bed rise for the debris to spill out.
Solution:
Let x be the height of the front of the bed rise needed to be raised for the debris to spill out. We can find x using trigonometric identities. That is:
sin θ = opposite / hypotenuse
Using trigonometric identities, we can get that:
sin(30) = x / 12
This gives:
0.5 = x / 12
Cross multiplying the terms to get:
x = 12 * 0.5
x = 6 ft
Therefore the front of the bed rise must be raised 6 ft for the debris to spill out.
Answer:
Step-by-step explanation:
Area = 84sqaure inch
Width = 2 1/2inch = 5/2inch
Area = l × width
So: 84 = l × 5/2
168 = 5l
l =>168/5
l = 33 3/5