1) X^5Y/X^5= X^0Y= Y (Anything raise to power of 0 is 1)
2)(3mn)^1= 3mn
3)m^-4xM^5= M^-4+5= M
4) 5^-4x5^4= 5^-4+4= 5^0= 1
5) nm^3/n^3m= n^1-3m^3-1= n^-2m^2= m^2/n^2
6) (-2)^0= 1
therefore number 4 and 6 has the value of 1
Answer:
I DO LOL
Step-by-step explanation:
sorry for caps
Answer:
B). The left and right-hand edges of the box will be approximately equal distances from the median.
Step-by-step explanation:
The 'symmetry' si described as a 'satisfying arrangement of a balanced distribution of the elements of the whole' while a 'symmetry group' is characterized as a group whose elements are all the transformations under which a given object remains invariant and whose group operation is function composition.
As per the given conditions, <u>the second statement asserts a true claim regarding the keeping of edges of left, as well as, right-hand side's boxes at equal intervals from the median</u>. This will help in making the arrangement fulfilling while keeping a little scope for the variation. Thus, <u>option B</u> is the correct answer.
Answer:
Step-by-step explanation:
0.25230452
Answer: -6
Step-by-step explanation: use the distributive property and distribute the two to the n and 5. that’ll leave you with 2n+10=-2. then you subtract 10 from both sides and get left with 2n=-12. and then to get the exponent by itself, you divide by two on both sides. and then you’re left with n=-6 :)