I think it's 12 because here it comes out
Answer: No triangle can be drawn.
Step-by-step explanation:
According to the triangle Inequality theorem, the sum of any two sides of a triangle must be greater than the third side.
If the measurements of triangle taken as 2 cm, 3 cm, and 5 cm , no triangle can be formed because 2+3=5 which does not satisfy the triangle inequality.
If the measurements of triangle taken as 2 cm, 3 cm, and 8 cm , no triangle can be formed because 2+3=5<8 which does not satisfy the triangle inequality.
Hence, No triangle can be drawn.
The answer is <span>x < −8
</span>
<span>3 − (2x − 5) < −4(x + 2)
3 - 2x + 5 < -4x - 8
8 - 2x < -4x - 8
4x - 2x < -8 - 8
2x < -16
x < -8</span>
Answer:
is the correct answer.
Step-by-step explanation:
Given:
![$\frac{5}{2+\sqrt{6}}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B5%7D%7B2%2B%5Csqrt%7B6%7D%7D%24)
To find:
if given term is written as following:
![$\frac{A\sqrt{B}+C}{D}$](https://tex.z-dn.net/?f=%24%5Cfrac%7BA%5Csqrt%7BB%7D%2BC%7D%7BD%7D%24)
<u>Solution:</u>
We can see that the resulting expression does not contain anything under
(square root) so we need to rationalize the denominator to remove the square root from denominator.
The rule to rationalize is:
Any term having square root term in the denominator, multiply and divide with the expression by changing the sign of square root term of the denominator.
Applying this rule to rationalize the given expression:
![\dfrac{5}{2+\sqrt{6}} \times \dfrac{2-\sqrt6}{2-\sqrt6}\\\Rightarrow \dfrac{5 \times (2-\sqrt6)}{(2+\sqrt{6}) \times (2-\sqrt6)} \\\Rightarrow \dfrac{10-5\sqrt6}{2^2-(\sqrt6)^2}\ \ \ \ \ (\because \bold{(a+b)(a-b)=a^2-b^2})\\\Rightarrow \dfrac{10-5\sqrt6}{4-6}\\\Rightarrow \dfrac{10-5\sqrt6}{-2}\\\Rightarrow \dfrac{-5\sqrt6+10}{-2}\\\Rightarrow \dfrac{5\sqrt6-10}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B5%7D%7B2%2B%5Csqrt%7B6%7D%7D%20%5Ctimes%20%5Cdfrac%7B2-%5Csqrt6%7D%7B2-%5Csqrt6%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B5%20%5Ctimes%20%282-%5Csqrt6%29%7D%7B%282%2B%5Csqrt%7B6%7D%29%20%5Ctimes%20%282-%5Csqrt6%29%7D%20%5C%5C%5CRightarrow%20%5Cdfrac%7B10-5%5Csqrt6%7D%7B2%5E2-%28%5Csqrt6%29%5E2%7D%5C%20%5C%20%5C%20%5C%20%5C%20%20%20%28%5Cbecause%20%5Cbold%7B%28a%2Bb%29%28a-b%29%3Da%5E2-b%5E2%7D%29%5C%5C%5CRightarrow%20%5Cdfrac%7B10-5%5Csqrt6%7D%7B4-6%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B10-5%5Csqrt6%7D%7B-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B-5%5Csqrt6%2B10%7D%7B-2%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B5%5Csqrt6-10%7D%7B2%7D)
Comparing the above expression with:
![$\frac{A\sqrt{B}+C}{D}$](https://tex.z-dn.net/?f=%24%5Cfrac%7BA%5Csqrt%7BB%7D%2BC%7D%7BD%7D%24)
A = 5, B = 6 (Not divisible by square of any prime)
C = -10
D = 2 (positive)
GCD of A, C and D is 1.
So, ![A +B+C+D = 5+6-10+2 = \bold3](https://tex.z-dn.net/?f=A%20%2BB%2BC%2BD%20%3D%205%2B6-10%2B2%20%3D%20%5Cbold3)