X = 2
You take 7 and subtract 5 to get x
<u>Prove that:</u>
![\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5Csf%5C%3A%5C%3A%5Cleft%28%5Cdfrac%7Bb%5E2-c%5E2%7D%7Ba%7D%5Cright%29%5Ccos%20A%2B%5Cleft%28%5Cdfrac%7Bc%5E2-a%5E2%7D%7Bb%7D%5Cright%29%5Ccos%20B%20%2B%5Cleft%28%5Cdfrac%7Ba%5E2-b%5E2%7D%7Bc%7D%5Cright%29%5Ccos%20C%3D0)
<u>Proof: </u>
We know that, by Law of Cosines,
<u>Taking</u><u> </u><u>LHS</u>
<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>
![\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)](https://tex.z-dn.net/?f=%5Clongmapsto%5Cleft%28%5Cdfrac%7Bb%5E2-c%5E2%7D%7Ba%7D%5Cright%29%5Cleft%28%5Cdfrac%7Bb%5E2%2Bc%5E2-a%5E2%7D%7B2bc%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7Bc%5E2-a%5E2%7D%7Bb%7D%5Cright%29%5Cleft%28%5Cdfrac%7Bc%5E2%2Ba%5E2-b%5E2%7D%7B2ca%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7Ba%5E2-b%5E2%7D%7Bc%7D%5Cright%29%5Cleft%28%5Cdfrac%7Ba%5E2%2Bb%5E2-c%5E2%7D%7B2ab%7D%5Cright%29%20)
![\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)](https://tex.z-dn.net/?f=%5Clongmapsto%5Cleft%28%5Cdfrac%7B%28b%5E4-c%5E4%29-%28a%5E2b%5E2-a%5E2c%5E2%29%7D%7B2abc%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7B%28c%5E4-a%5E4%29-%28b%5E2c%5E2-a%5E2b%5E2%29%7D%7B2abc%7D%5Cright%29%2B%5Cleft%28%5Cdfrac%7B%28a%5E4-b%5E4%29-%28a%5E2c%5E2-b%5E2c%5E2%29%7D%7B2abc%7D%5Cright%29)
![\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}](https://tex.z-dn.net/?f=%5Clongmapsto%5Cdfrac%7Bb%5E4-c%5E4-a%5E2b%5E2%2Ba%5E2c%5E2%7D%7B2abc%7D%2B%5Cdfrac%7Bc%5E4-a%5E4-b%5E2c%5E2%2Ba%5E2b%5E2%7D%7B2abc%7D%2B%5Cdfrac%7Ba%5E4-b%5E4-a%5E2c%5E2%2Bb%5E2c%5E2%7D%7B2abc%7D)
<em>On combining the fractions,</em>
![\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}](https://tex.z-dn.net/?f=%5Clongmapsto%5Cdfrac%7Bb%5E4-c%5E4-a%5E2b%5E2%2Ba%5E2c%5E2%2Bc%5E4-a%5E4-b%5E2c%5E2%2Ba%5E2b%5E2%2Ba%5E4-b%5E4-a%5E2c%5E2%2Bb%5E2c%5E2%7D%7B2abc%7D)
<em>Regrouping the terms,</em>
![\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}](https://tex.z-dn.net/?f=%5Clongmapsto%5Cdfrac%7B%28a%5E4-a%5E4%29%2B%28b%5E4-b%5E4%29%2B%28c%5E4-c%5E4%29%2B%28a%5E2b%5E2-a%5E2b%5E2%29%2B%28b%5E2c%5E2-b%5E2c%5E2%29%2B%28a%5E2c%5E2-a%5E2c%5E2%29%7D%7B2abc%7D)
![\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}](https://tex.z-dn.net/?f=%5Clongmapsto%5Cdfrac%7B%280%29%2B%280%29%2B%280%29%2B%280%29%2B%280%29%2B%280%29%7D%7B2abc%7D)
![\longmapsto\bf 0=RHS](https://tex.z-dn.net/?f=%5Clongmapsto%5Cbf%200%3DRHS)
LHS = RHS proved.
Answer:
Correct answer: a = 3 · 4⁽ⁿ ⁻ ¹⁾
Step-by-step explanation:
Given:
Geometric sequence 3, 12, 48, 192, .....
First term a₁ = 3
Second term a₂ = 12
Third term a₃ = 48
Common ratio or quotient:
q = a₂ / a₁ = a₃ / a₂ = 12 / 3 = 48 / 12 = 4
q = 4
First term a₁ = 3
Second term a₂ = a · q
Third term a₃ = a₂ · q = a₁ · q²
Fourth term a₄ = a₃ · q = a₁ · q³
......................................................
n- th term aₙ = a₁ · q⁽ⁿ ⁻ ¹⁾
In this case aₙ = 3 · 4⁽ⁿ ⁻ ¹⁾
God is with you!!!
Answer:
2/3
Step-by-step explanation:
Answer:
SL = KR
Step-by-step explanation:
We're given two different lengths, both of which are on other sides of KL. Now, since these two givens are congruent, and we're overlapping it over a new line, the lines SL and KR are going to be congruent.