(1,9) (-3,5)
Midpoint = (1+{-3}/2) (9+5/2)
(1-3/2) (14/2)
(-2/2) (7)
(-1) (7)
Midpoint Answer = (-1, 7)
Y=mx+b
mx=b-y
÷x. ÷x
m=b-y/x
Answer:
Hello There!!
Step-by-step explanation:
Firstly, find out how many students there are in 7 batches. You can do this by timesing 48 by 7 to get 336. So there are 336 students in 7 batches. Each student does 24 hours of community service in a year so do 336 times by 24 which is 8,064. Therefore,the college contributes 8,064 hours of community service in a year.
hope this helps,have a great day!!
~Pinky~
Answer:
Solution to determine whether each of these sets is countable or uncountable
Step-by-step explanation:
If A is countable then there exists an injective mapping f : A → Z+ which, for any S ⊆ A gives an injective mapping g : S → Z+ thereby establishing that S is countable. The contrapositive of this is: if a set is not countable then any superset is not countable.
(a) The rational numbers are countable (done in class) and this is a subset of the rational. Hence this set is also countable.
(b) this set is not countable. For contradiction suppose the elements of this set in (0,1) are enumerable. As in the diagonalization argument done in class we construct a number, r, in (0,1) whose decimal representation has as its i th digit (after the decimal) a digit different from the i th digit (after the decimal) of the i th number in the enumeration. Note that r can be constructed so that it does not have a 0 in its representation. Further, by construction r is different from all the other numbers in the enumeration thus yielding a contradiction
![x= \sqrt[3]{2*2*2*2}](https://tex.z-dn.net/?f=x%3D%20%5Csqrt%5B3%5D%7B2%2A2%2A2%2A2%7D%20)
prime factorization of 16![x=2 \sqrt[3]{2}](https://tex.z-dn.net/?f=x%3D2%20%5Csqrt%5B3%5D%7B2%7D%20)
a set of 3 comes out as one number...i.e. 2