Answer:
2/5
Step-by-step explanation:
Reflection across the diagonal y=x.
180 degree clockwise rotation about the origin.
The ratio between the lengths of 2 rectangles is the same between their perimeters
![l_1\colon l_2=P_1\colon P_2](https://tex.z-dn.net/?f=l_1%5Ccolon%20l_2%3DP_1%5Ccolon%20P_2)
Since the ratio between the length of rectangle A and the length of rectangle B is 7: 3, then
The ratio between the perimeter of rectangle A to the perimeter of rectangle B is 7: 3 too
![\begin{gathered} l_A\colon l_B=7\colon3 \\ P_A\colon P_B=7\colon3 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20l_A%5Ccolon%20l_B%3D7%5Ccolon3%20%5C%5C%20P_A%5Ccolon%20P_B%3D7%5Ccolon3%20%5Cend%7Bgathered%7D)
Since the perimeter of rectangle A is 540, then
![540\colon P_B=7\colon3](https://tex.z-dn.net/?f=540%5Ccolon%20P_B%3D7%5Ccolon3)
We will write them as a fraction
![\frac{540}{P_B}=\frac{7}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B540%7D%7BP_B%7D%3D%5Cfrac%7B7%7D%7B3%7D)
By using the cross multiplication
![\begin{gathered} P_B\times7=540\times3 \\ 7P_B=1620 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P_B%5Ctimes7%3D540%5Ctimes3%20%5C%5C%207P_B%3D1620%20%5Cend%7Bgathered%7D)
Divide both sides by 7
![\begin{gathered} \frac{7P_B}{7}=\frac{1620}{7} \\ P_B=231.4285714 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7B7P_B%7D%7B7%7D%3D%5Cfrac%7B1620%7D%7B7%7D%20%5C%5C%20P_B%3D231.4285714%20%5Cend%7Bgathered%7D)
Round it to the nearest tenth
![P_B=231.4\text{ inches}](https://tex.z-dn.net/?f=P_B%3D231.4%5Ctext%7B%20inches%7D)
The perimeter of the smaller rectangle B is 231.4 inches
Answer:
![x=0, \\x=\pi](https://tex.z-dn.net/?f=x%3D0%2C%20%5C%5Cx%3D%5Cpi)
Step-by-step explanation:
Recall the trigonometric identity
.
Therefore, given
, rewrite the left side of the equation:
![2\sin x\cos x=2\sin x](https://tex.z-dn.net/?f=2%5Csin%20x%5Ccos%20x%3D2%5Csin%20x)
Subtract
from both sides:
![2\sin x\cos x-2\sin x=0](https://tex.z-dn.net/?f=2%5Csin%20x%5Ccos%20x-2%5Csin%20x%3D0)
Factor out
from both terms on the left:
![2\sin x(\cos x-1)=0](https://tex.z-dn.net/?f=2%5Csin%20x%28%5Ccos%20x-1%29%3D0)
We now have two cases:
![\begin{cases}2\sin x=0, x=k\pi\text{ for }k\in \mathbb{Z}\\\cos x-1=0,x=k2\pi\text{ for }k\in \mathbb{Z}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D2%5Csin%20x%3D0%2C%20x%3Dk%5Cpi%5Ctext%7B%20for%20%7Dk%5Cin%20%5Cmathbb%7BZ%7D%5C%5C%5Ccos%20x-1%3D0%2Cx%3Dk2%5Cpi%5Ctext%7B%20for%20%7Dk%5Cin%20%5Cmathbb%7BZ%7D%5Cend%7Bcases%7D)
Since the problem stipulates that
is in the interval
, we have:
![\text{For }x\in [0, 2\pi):\\\sin x=0\rightarrow \boxed{x=0, x=\pi}\\\cos x-1=0\rightarrow \boxed{x=0}](https://tex.z-dn.net/?f=%5Ctext%7BFor%20%7Dx%5Cin%20%5B0%2C%202%5Cpi%29%3A%5C%5C%5Csin%20x%3D0%5Crightarrow%20%5Cboxed%7Bx%3D0%2C%20x%3D%5Cpi%7D%5C%5C%5Ccos%20x-1%3D0%5Crightarrow%20%5Cboxed%7Bx%3D0%7D)
Recall that square brackets mean inclusive and parentheses mean exclusive. Therefore,
.
Try this:
21/7=3
multiply 3x2
x=6
y=21 x=6
or...
7/2=21/x
cross multiply to get 7x=42
divide by 7 and x=6