It’s not that hard all u need to do is when u multiply it just multiply it. For example what’s 2/3x4/3=8/3 and that is just 8/3 but when the denominator is different u just multiply it it’s not that hard
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: cos 330 = ![\frac{\sqrt3}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt3%7D%7B2%7D)
Use the Double-Angle Identity: cos 2A = 2 cos² A - 1
![\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}](https://tex.z-dn.net/?f=%5Ctext%7BScratchwork%3A%7D%5Cquad%20%5Cbigg%28%5Cdfrac%7B%5Csqrt3%20%2B%202%7D%7B2%5Csqrt2%7D%5Cbigg%29%5E2%20%3D%20%5Cdfrac%7B2%5Csqrt3%20%2B%204%7D%7B8%7D)
Proof LHS → RHS:
LHS cos 165
Double-Angle: cos (2 · 165) = 2 cos² 165 - 1
⇒ cos 330 = 2 cos² 165 - 1
⇒ 2 cos² 165 = cos 330 + 1
Given: ![2 \cos^2 165 = \dfrac{\sqrt3}{2} + 1](https://tex.z-dn.net/?f=2%20%5Ccos%5E2%20165%20%20%3D%20%5Cdfrac%7B%5Csqrt3%7D%7B2%7D%20%2B%201)
![\rightarrow 2 \cos^2 165 = \dfrac{\sqrt3}{2} + \dfrac{2}{2}](https://tex.z-dn.net/?f=%5Crightarrow%202%20%5Ccos%5E2%20165%20%20%3D%20%5Cdfrac%7B%5Csqrt3%7D%7B2%7D%20%2B%20%5Cdfrac%7B2%7D%7B2%7D)
Divide by 2: ![\cos^2 165 = \dfrac{\sqrt3+2}{4}](https://tex.z-dn.net/?f=%5Ccos%5E2%20165%20%20%3D%20%5Cdfrac%7B%5Csqrt3%2B2%7D%7B4%7D)
![\rightarrow \cos^2 165 = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}](https://tex.z-dn.net/?f=%5Crightarrow%20%5Ccos%5E2%20165%20%20%3D%20%5Cbigg%28%5Cdfrac%7B2%7D%7B2%7D%5Cbigg%29%5Cdfrac%7B%5Csqrt3%2B2%7D%7B4%7D)
Square root: ![\sqrt{\cos^2 165} = \sqrt{\dfrac{4+2\sqrt3}{8}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Ccos%5E2%20165%7D%20%20%3D%20%5Csqrt%7B%5Cdfrac%7B4%2B2%5Csqrt3%7D%7B8%7D%7D)
Scratchwork: ![\cos^2 165 = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2](https://tex.z-dn.net/?f=%5Ccos%5E2%20165%20%20%3D%20%5Cbigg%28%5Cdfrac%7B%5Csqrt3%2B1%7D%7B2%5Csqrt2%7D%5Cbigg%29%5E2)
![\rightarrow \cos 165 = \pm \dfrac{\sqrt3+1}{2\sqrt2}](https://tex.z-dn.net/?f=%5Crightarrow%20%5Ccos%20165%20%20%3D%20%5Cpm%20%5Cdfrac%7B%5Csqrt3%2B1%7D%7B2%5Csqrt2%7D)
Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE
![\rightarrow \cos 165 = - \dfrac{\sqrt3+1}{2\sqrt2}](https://tex.z-dn.net/?f=%5Crightarrow%20%5Ccos%20165%20%20%3D%20-%20%5Cdfrac%7B%5Csqrt3%2B1%7D%7B2%5Csqrt2%7D)
LHS = RHS ![\checkmark](https://tex.z-dn.net/?f=%5Ccheckmark)
Answer:
The answer is below
Step-by-step explanation:
Given that Triangle PTS has all sides with equal lengths, it is considered is an equilateral triangle.
Similarly, with < TPS and <TSP aare both equilateral triangle because they have equal length.
Therefore, ∆ PQS and ∆ SRP are congruent based on side-angle-side theorem.
Similarly, given the fact that their congruent sides of congruent triangles are congruent, both Segment PQ is congruent to segment SR.
Also, Segment PR is congruent to segment SQ as a result of congruent sides of the congruent triangles are congruent.
Lastly, the ∆ PQR and ∆ SRQ are congruent based on the side-side-side theorem.
Answer: If my calculations are correct, it'd be 160% percent decrease
Step-by-step explanation:
you subtract 640 and 480, 640%-480%= 160%
It’s very easy lol it’s B