well, let's first find the slope of "this" line hmmmmm what would that be?
![\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-(-6)}{-2-1}\implies \cfrac{6+6}{-2-1} \\\\\\ \cfrac{12}{-3}\implies -4](https://tex.z-dn.net/?f=%20%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-6%7D%29%5Cqquad%20%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%20%5C%5C%5C%5C%5C%5C%20slope%20%3D%20%20m%5Cimplies%20%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B6-%28-6%29%7D%7B-2-1%7D%5Cimplies%20%5Ccfrac%7B6%2B6%7D%7B-2-1%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B12%7D%7B-3%7D%5Cimplies%20-4%20)
![\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-4\implies -\cfrac{4}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{1}{4}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{4}}}\implies \cfrac{1}{4}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bperpendicular%20lines%20have%20%5Cunderline%7Bnegative%20reciprocal%7D%20slopes%7D%7D%20%7B%5Cstackrel%7Bslope%7D%7B-4%5Cimplies%20-%5Ccfrac%7B4%7D%7B1%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7Breciprocal%7D%7B-%5Ccfrac%7B1%7D%7B4%7D%7D%5Cqquad%20%5Cstackrel%7Bnegative~reciprocal%7D%7B%2B%5Ccfrac%7B1%7D%7B4%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B4%7D%20)
Answer:
√5 /2- 1 or 0.1180 to the nearest 10,000th.
Step-by-step explanation:
3 sin A = 2
sin A = 2/3
The adjacent ( to < A) leg of the right triangle = √(3^2 - 2^2) = √5.
So cot A = adjacent / opposite = √5 / 2
and cot A - 1 = √5/2 - 1.
Use the vertical line rule to test.
Imagine you are drawing lines parallel to the y axis, if any of your lines touches the graph twice, the graph does not represent a function. otherwise, it does.
Answer:
15.) 3 5
27.) 3 9
56.) 2 4
16.) 2 4
28.) 2 4
57.) 3
102.) 2 3 6
268.) 2 4
4518.) 2 3 6 9
93.) 3
144.) 2 3 4 6 9
256.) 2 4
75.) 3 5
450.) 2 3 5 6 9 10
70.) 2 5 10
Step-by-step explanation:
15.) 3 x 5
27.) 3 x 9
56.) 2 x 28 || 4 x 14
16.) 2 x 8 || 4 x 4
28.) 2 x 14 || 4 x 7
57.) 3 x 19
102.) 2 x 51 || 3 x 34 || 6 x 17
268.) 2 x 134 || 4 x 67
4518.) 2 x 2259 || 3 x 1506 || 6 x 753 || 9 x 502
93.) 3 x 31
144.) 2 x 72 || 3 x 48 || 4 x 36 || 6 x 24 || 9 x 16
256.) 2 x 128 || 4 x 64
75.) 3 x 25 || 5 x 15
450.) 2 x 225 || 3 x 150 || 5 x 90 || 6 x 75 || 9 x 50 || 10 x 45
70.) 2 x 35 || 5 x 14 || 7 x 10
To solve this problem, we can use cross multiplication:
First, we can to convert inches into feet so...
5 inches = 0.416667 ft
7 inches = 0.583333 ft
(0.416667 ft/ 0.583333 ft) = (12 ft/ x ft)
x = 16.8 ft
So the actual width is 16.8 ft.
Hope this helps!