function RxBrioUtils.combineLatest(b)assert(type(b)=="\116\097\098\108\101","\066\097\100 \111\098\115\101\114\118\097\098\108\101\115")warn("\091\082\120\066\114\105\111\085\116\105\108\115\046\099\111\109\098\105\110\101\076\097\116\101\115\116\093 - \068\101\112\114\101\099\097\116\101\100 \115\105\110\099\101 \051\046\054\046\048\046 \085\115\101 \082\120\066\114\105\111\085\116\105\108\115\046\102\108\097\116\067\111\109\098\105\110\101\076\097\116\101\115\116")return(Rx.combineLatest(b)):Pipe({Rx.map(BrioUtils.flatten);RxBrioUtils.onlyLastBrioSurvives()})end
--[=[
	Flattens all the brios in one brio and combines them. Note that this method leads to
	gaps in the lifetime of the brio.

	@deprecated 3.6.0 -- This method does not wrap the resulting value in a Brio, which can sometimes lead to leaks.
	@param observables { [any]: Observable<Brio<T>> | Observable<T> | T }
	@return Observable<Brio<{ [any]: T }>>
]=]
function RxBrioUtils.combineLatest(observables)
	assert(type(observables) == "table", "Bad observables")

	warn("[RxBrioUtils.combineLatest] - Deprecated since 3.6.0. Use RxBrioUtils.flatCombineLatest")

	return Rx.combineLatest(observables):Pipe({
		Rx.map(BrioUtils.flatten),
		RxBrioUtils.onlyLastBrioSurvives(),
	})
end
