When you multiply them it equals a constant.
Answer:
![x = 2.5](https://tex.z-dn.net/?f=x%20%3D%202.5)
Step-by-step explanation:
Given
![AC = 2x - 1](https://tex.z-dn.net/?f=AC%20%3D%202x%20-%201)
![AB = 2x + 1](https://tex.z-dn.net/?f=AB%20%3D%202x%20%2B%201)
![BC = 2\sqrt{7](https://tex.z-dn.net/?f=BC%20%3D%202%5Csqrt%7B7)
![\angle A = 60^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20A%20%3D%2060%5E%7B%5Ccirc%7D)
Required
Find x
To find x, we make use of cosine formula which states:
![a^2 = b^2 + c^2 - 2bcCosA](https://tex.z-dn.net/?f=a%5E2%20%3D%20b%5E2%20%2B%20c%5E2%20-%202bcCosA)
In this case:
![BC^2 = AB^2 + AC^2 - 2*AB*BC*CosA](https://tex.z-dn.net/?f=BC%5E2%20%3D%20AB%5E2%20%2B%20AC%5E2%20-%202%2AAB%2ABC%2ACosA)
Substitute values
![(2\sqrt{7})^2 = (2x + 1)^2 + (2x-1)^2 - 2*(2x + 1)*(2x - 1)*Cos(60^{\circ})](https://tex.z-dn.net/?f=%282%5Csqrt%7B7%7D%29%5E2%20%3D%20%282x%20%2B%201%29%5E2%20%2B%20%282x-1%29%5E2%20-%202%2A%282x%20%2B%201%29%2A%282x%20-%201%29%2ACos%2860%5E%7B%5Ccirc%7D%29)
Evaluate all exponents
![4*7 = (4x^2 + 4x + 1)+(4x^2 - 4x + 1) - 2(4x^2 - 1)*cos(60^{\circ})](https://tex.z-dn.net/?f=4%2A7%20%3D%20%284x%5E2%20%2B%204x%20%2B%201%29%2B%284x%5E2%20-%204x%20%2B%201%29%20-%202%284x%5E2%20-%201%29%2Acos%2860%5E%7B%5Ccirc%7D%29)
Open brackets
![28 = 4x^2 + 4x + 1+4x^2 - 4x + 1 - (8x^2 - 2)*cos(60^{\circ})](https://tex.z-dn.net/?f=28%20%3D%204x%5E2%20%2B%204x%20%2B%201%2B4x%5E2%20-%204x%20%2B%201%20-%20%288x%5E2%20-%202%29%2Acos%2860%5E%7B%5Ccirc%7D%29)
Collect Like Terms
![28 = 4x^2 +4x^2+ 4x - 4x+ 1 + 1 - (8x^2 - 2)*cos(60^{\circ})](https://tex.z-dn.net/?f=28%20%3D%204x%5E2%20%2B4x%5E2%2B%204x%20-%204x%2B%201%20%20%2B%201%20-%20%288x%5E2%20-%202%29%2Acos%2860%5E%7B%5Ccirc%7D%29)
![28 = 8x^2+ 2 - (8x^2 - 2)*cos(60^{\circ})](https://tex.z-dn.net/?f=28%20%3D%208x%5E2%2B%202%20-%20%288x%5E2%20-%202%29%2Acos%2860%5E%7B%5Ccirc%7D%29)
Substitute 0.5 for cos(60)
![28 = 8x^2+ 2 - (8x^2 - 2)*0.5](https://tex.z-dn.net/?f=28%20%3D%208x%5E2%2B%202%20-%20%288x%5E2%20-%202%29%2A0.5)
Open bracket
![28 = 8x^2+ 2 - 4x^2 + 1](https://tex.z-dn.net/?f=28%20%3D%208x%5E2%2B%202%20-%204x%5E2%20%2B%201)
Collect Like Terms
![8x^2- 4x^2 + 2 + 1 - 28 = 0](https://tex.z-dn.net/?f=8x%5E2-%204x%5E2%20%2B%202%20%2B%201%20-%2028%20%3D%200)
![4x^2 -25 = 0](https://tex.z-dn.net/?f=4x%5E2%20%20-25%20%3D%200)
Collect Like Terms
![4x^2 =25](https://tex.z-dn.net/?f=4x%5E2%20%20%3D25)
Divide both sides by 4
![x^2 =\frac{25}{4}](https://tex.z-dn.net/?f=x%5E2%20%20%3D%5Cfrac%7B25%7D%7B4%7D)
![x^2 =6.25](https://tex.z-dn.net/?f=x%5E2%20%20%3D6.25)
Take positive square root of both sides
![x = \sqrt{6.25](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%7B6.25)
![x = 2.5](https://tex.z-dn.net/?f=x%20%3D%202.5)
The answer is 6 have a nice day
Answer:
congrats, you just put a question
anyways have a nice day and thanks for the pointssss
Answer:
the greatest possible value for x is 6
Explanation:
Assume that the sides of the triangle are:
a = x
b = 3x
c = 20
We are given that c is the longest side.
For the triangle to be obtuse:
c^2 > a^2 + b^2
Substitute with the values of a, b and c and solve for x as follows:
c^2 > a^2 + b^2
(20)^2 > (x)^2 + (3x)^2
400 > x^2 + 9x^2
400 > 10 x^2
40 > x^2
Now, we will get the zeros of the function. This means that we will solve for x^2 = 40 as follows:
x^2 = 40
x = + or - √40
either x = 6.32
Or x = -6.32
Since we want x^2 < 40, therefore, the greatest possible value for x to satisfy this condition is 6
Hope this helps :)