<span>The Java code below will ask for two integers and display the sum. If a non-integer is submitted the code will ask again. The break is used to exit the while true loop indicating that no-errors had occurred and two numbers were added.
System.out.print("Please insert two integers and this will display the sum.");
int numOne;
int numTwo;
while (True) {
try{
System.out.print("Integer Number One? ");
numOne = input.nextInt();
System.out.print("Integer Number Two? ");
numTwo = input.nextInt();
System.out.print("The Sum Is: " + (numOne + numTwo));
break;
}
catch (InputMismatchException e) {
System.out.print("please enter an integer .");
}
}</span>
C on that one which is the correct answer
Option A
is one way to determine the factors of
by grouping
<em><u>Solution:</u></em>
Factoring by grouping means that you will group terms with common factors before factoring
<em><u>Given expression is:</u></em>

Group the first two terms together and then the last two terms together.

We can see that
is common in first two terms
And 3 is common in last two terms
Factor them out


Thus option A is correct
Answer:
212m
Step-by-step explanation:
The set up will be equivalent to a right angled triangle where the height is the opposite side facing the 45° angle directly. The length of the rope will be the slant side which is the hypotenuse.
Using the SOH, CAH, TOA trigonometry identity to solve for the length of the rope;
Since we have the angle theta = 45° and opposite = 150m
According to SOH;
Sin theta = opposite/hypotenuse.
Sin45° = 150/hyp
hyp = 150/sin45°
hyp = 150/(1/√2)
hyp = 150×√2
hyp = 150√2 m
hyp = 212.13m
Hence the length of the rope for the kite sail, in order to pull the ship at an angle of 45° and be at a vertical height of 150 m is approximately 212m
1. A straight line segment can be drawn joining any two points.
2. Any straight line segment can be extended indefinitely in a straight line.
3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.
4. All right angles are congruent.
5. If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. This postulate is equivalent to what is known as the parallel postulate.