Find the inertia tensor for an equilateral triangle in the xy plane. Take the mass of the triangle to be M and the length of a side of the triangle to be b. Express your answer below as pure numbers in units of Mb^2. Place the origin on the midpoint of one side and set the y-axis to be along the symmetry axis.
Answer:
uyjytyhy
Step-by-step explanation:
tthtyi
tyujftyu
yijy
The second step is wrong. What should've been done is to find greatest common factor (gcf) of 1/6 and -2. This is because you cannont add together a number with a variable to a number without a variable. So get the variable by itself by subtracting 1/6 from both sides.
1/5x + 1/6 = -2
___— 1/6_— 1/6
____________
Turn -2 into a fraction and find the gcf of -2 and 6:
1/5x + 1/6 = -2
___— 1/6_— 1/6
____________
1/5x = -2/1 — 1/6 ——> 1/5x = -12/6 — 1/6
1/5x = -13/6
Then divide each side by 1/5 to get the variable by itself; remeber: when dividing a fraction by a fraction, you multiply by the reciprocal.
5/1 • 1/5x = -13/6 • 5/1
x = 65/6
Then, simplify
65/6} 10.83 or 10 83/100
Answer:
S’ (-5,-3)
T’ (4,-6)
U’ (-3,-7)
Step-by-step explanation:
Basically, we want to apply the given transformation rule of the axes to get the coordinates of the images
The rule is that we subtract 1 from the x-coordinates and also subtract 5 from the y-coordinates
Thus, we proceed as follows;
S’ = (-4-1, 2-5) = (-5,-3)
T’ = (5-1, -1-5) = (4,-6)
U’ = (-2-1,-2-5) = (-3,-7)
Answer:
91
Step-by-step explanation:
7(10+3)
7(10)+7(3)
70+21
91
