Start by decomposing the number inside the root into primes
Then group the terms into cubes if possible
rewrite the root
![\sqrt[3]{80}=\sqrt[3]{10\cdot2^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B80%7D%3D%5Csqrt%5B3%5D%7B10%5Ccdot2%5E3%7D)
then cancel the terms that are cubes and bring them out of the root
![\sqrt[3]{80}=2\sqrt[3]{10}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B80%7D%3D2%5Csqrt%5B3%5D%7B10%7D)
Answer:
The answer is "MS and QS".
Step-by-step explanation:
Given ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. we have to prove that ΔMNS ≅ ΔQNS.
As NR and MQ bisect each other at S
⇒ segments MS and SQ are therefore congruent by the definition of bisector i.e MS=SQ
In ΔMNS and ΔQNS
MN=QN (∵ MNQ is isosceles triangle)
∠NMS=∠NQS (∵ MNQ is isosceles triangle)
MS=SQ (Given)
By SAS rule, ΔMNS ≅ ΔQNS.
Hence, segments MS and SQ are therefore congruent by the definition of bisector.
The correct option is MS and QS
X is equal 3. to do this, you do reverse of order of operations. So this means that you would to the 'times 2' first by dividing 4 by 2 to make it 2. then, in the parenthesis, since there is a 'minus 1', therefore you are going to add 1+2 which equals three. <span />
We can’t see the bar model
