it would be 40000000 because there there are 7 zeroes and then you and then you multiply the two twos together and then put it in the front.
If you're only provided with the lengths of a triangle, and you're asked to determine whether or not the triangle is right or not, you'll need to rely on the Pythagorean Theorem to help you out. In case you're rusty on it, the Pythagorean Theorem defines a relationship between the <em>legs</em> of a right triangle and its <em>hypotenuse</em>, the side opposite its right angle. That relationship is a² + b² = c², where a and b are the legs of the triangle, and c is its hypotenuse. To see if our triangle fits that requirement, we'll have to substitute its lengths into the equation.
How do we determine which length is the hypotenuse, though? Knowledge that the hypotenuse is always the longest length of a right triangle helps here, as we can clearly observe that 8.6 is the longest we've been given for this problem. The order we pick the legs in doesn't matter, since addition is commutative, and we'll get the same result regardless of the order we're adding a and b.
So, substituting our values in, we have:
(2.6)² + (8.1)² = (8.6)²
Performing the necessary calculations, we have:
6.76 + 65.61 = 73.96
72.37 ≠ 73.96
Failing this, we know that our triangle cannot be right, but we <em>do </em>know that 72.37 < 73.96, which tells us something about what kind of triangle it is. Imagine taking a regular right triangle and stretching its hypotenuse, keeping the legs a and b the same length. This has the fact of <em>increasing the angle between a and b</em>. Since the angle was already 90°, and it's only increased since then, we know that the triangle has to be <em>obtuse</em>, which is to say: yes, there's an angle in it larger than 90°.
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:
∠FJG= 135
Step-by-step explanation:
∠FJG + ∠GJH = 168 so
(6x - 15) + (x + 8) = 168
7x - 7 = 168
7x = 175
x = 25
plug it in ∠FJG
6(25) - 15 = 135
y = -0.65 (6) + 7.34
y = 3.44
but the dot plot showed when x = 6, y = 4
so 4 - 3.44 = 0.56
Answer
0.56
