Answer:
The second one:
int sum = 0; for (int i = 0; i < values.length; i++) { if ((values[i] % 2) == 0) { sum += values[i]; } }
A dial-up access connection supports a speed up to <u>5</u><u>6kbps</u>, whereas an ISDN line goes up to 1.9Mbps. A DSL connection can support a maximum of 20<u>mbps</u>. Internet access speed through cable TV is capable of up to 1000mbps.
Dial-up connection is by far the slowest of all known internet connections. The maximum speeds that can be supported by this connection are about 56Kbps. Use of this type of connection requires a separate phone line. In an ISDN type of connection, the maximum speeds depend on where in the world you are. The highest to have ever been recorded is an ISDN E1 line that has a combined data rate of 1.9 Mbit/s. Speeds transmitted through DSL are generally the same ones that are transmitted through cable internet and satellite connection. You can expect DSL speeds of 512Kbps to a max of 20Mbps. The cable internet is a broadband internet access and the highest bit rates of cable internet can go be up to 1GBPS which is equivalent to 1000Mbps.
Answer:
The answer to this question can be given as:
Statement:
isQuadrilateral = (numberOfSides == 4) ? 1 : 0;
//check condition using ternary operator.
Explanation:
we know that both(Quadrilateral, numberOfSides) is already declared in the program. So the statement for check condition is (isQuadrilateral = (numberOfSides == 4) ? 1 : 0;). To check this condition we use the ternary operator. In this operator, we also check another condition. The syntax of ternary operator (condition ? value_if_true : value_if_false).In this statement on the lift side we use the variable for check condition and right side we check condition if the value is true it prints 1 else it will print 0.
Answer:
Check the explanation
Explanation:
We can utilize the above algorithm with a little in modification. If in each of the iteration, we discover a node with no inward edges, then we we’re expected succeed in creating a topological ordering.
If in a number of iteration, it becomes apparent that each of the node has a minimum of one inward edge, then there must be a presence of cycle in the graph.
So our algorithm in finding the cycle is this: continually follow an edge into the node we’re presently at (which is by choosing the first one on the adjacency list of inward edges to decrease the running time).
Since the entire node has an inward edge, we can do this continually or constantly until we revisit a node v for the first time.
The set of nodes that we will come across among these two successive visits is a cycle (which is traversed in the reverse direction).
