with the assumption this is a geometric sequence and thus it as an "r" common ratio, so we know that the 3rd term must be 12 * r and the 4th term must be 12 * r * r, so let's make a quick table
![\begin{array}{rll} term&value\\ \cline{1-2} a_2&12\\ a_3&12\cdot r\\\cline{1-2} a_4&(12\cdot r)r\\ &12r^2\\ &\frac{16}{3}\\\cline{1-2} a_5&12r^3\\ a_6&12r^4\\ a_7&12r^5\\ a_8&12r^6\\ a_9&12r^7 \end{array}\qquad \implies \begin{array}{llll} 12r^2=\cfrac{16}{3}\implies r^2=\cfrac{16}{12\cdot 3}\implies r^2=\cfrac{4}{9}\\\\\\ r=\sqrt{\cfrac{4}{9}}\implies r=\cfrac{\sqrt{4}}{\sqrt{9}}\implies r=\cfrac{2}{3} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brll%7D%20term%26value%5C%5C%20%5Ccline%7B1-2%7D%20a_2%2612%5C%5C%20a_3%2612%5Ccdot%20r%5C%5C%5Ccline%7B1-2%7D%20a_4%26%2812%5Ccdot%20r%29r%5C%5C%20%2612r%5E2%5C%5C%20%26%5Cfrac%7B16%7D%7B3%7D%5C%5C%5Ccline%7B1-2%7D%20a_5%2612r%5E3%5C%5C%20a_6%2612r%5E4%5C%5C%20a_7%2612r%5E5%5C%5C%20a_8%2612r%5E6%5C%5C%20a_9%2612r%5E7%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%2012r%5E2%3D%5Ccfrac%7B16%7D%7B3%7D%5Cimplies%20r%5E2%3D%5Ccfrac%7B16%7D%7B12%5Ccdot%203%7D%5Cimplies%20r%5E2%3D%5Ccfrac%7B4%7D%7B9%7D%5C%5C%5C%5C%5C%5C%20r%3D%5Csqrt%7B%5Ccfrac%7B4%7D%7B9%7D%7D%5Cimplies%20r%3D%5Ccfrac%7B%5Csqrt%7B4%7D%7D%7B%5Csqrt%7B9%7D%7D%5Cimplies%20r%3D%5Ccfrac%7B2%7D%7B3%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
Proven
Step-by-step explanation:
Let ABC be a triangle and D, E and F are midpoints of BC, CA and AB.
The sum of two sides of a triangle is greater than twice the median bisecting the third side,
Hence in ΔABD, AD is a median
⇒ AB + AC > 2(AD)
Hence, we get
BC + AC > 2 (CF
)
BC + AB > 2 (BE)
On adding the above inequalities, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2 (AD) + 2 (CD) + 2 (BE
)
2(AB + BC + AC) > 2(AD + BE + CF)
AB + BC + AC > AD + BE + CF - Proven
* Sorry I don't have any attachments, use your imagination!
Answer:
C Because divide 56/7 =8.
I think it’s the fourth one
It is prime, it can't be divided by anything, but itself and one.
