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We use two variables of type f 1 oat: main0 float inches, cm; tout << "Enter length in inches: "; tin >> inches; cm = 2.54*inches; tout << inches << ' inches = ' C-C cm C=C ' centimeters.\n"; )

(the repeated double quote at the end is correct)

Simply read the input into inches, 3.15

3. The distance from a point to a sphere is the external segment of the secant from the point through the center of the sphere.

Write a program to find the integer square root of a given number. That is the largest integer whose square is less than or equal to the given number. We use an exhaustive algorithm here: find &l the positive integers whose square is less than or equal to the given number; then the largest of those is the integer square root: main0 -t float x; tout << "Enter a positive number: "; tin >> x; for (int n = 1; n*n <= x; n++) ; // the null statement tout << "The integer square root of ' << x C-C ' is ' C-C n-l << endl;

5. Any point on the line which is the perpendicular bisector of a segment is equidistant from the ends of the segment.

We start with n = 1 and continue to increment n until n*n > x. When the for loop terminates, n is the smallest integer whose square is greater than x, so n- 1 is the integer square root of x. Notice the use of the null statement in the for loop. Everything that needs to be done in the loop is done within the control parts of the loop. But the semicolon is still necessary at the end of the loop.

5. Any point on the plane which is the perpendicular bisector of a segment is equidistant from the ends of the segment.

Write and run a program that directly implements the quotient operator / and the remainder operator % for the division of positive integers.

6. Any point on the line which is the bisector of the angle between two lines is equidistant from the sides of the angle.

The algorithm used here, applied to the fraction n/d, repeatedly subtracts the d from the n until n is less than d. At that point, the value of n will be the remainder, and the number q of iterations required to reach it will be the quotient: main0 { int n, d, q, r; tout << "Enter numerator: 'I; tin >> n; tout << "Enter denominator: 'I; tin >> d; for (q = 0, r = n; r > d; q++) r -= tout << n << I1 / I -c-c d x-c 'I = I' -c-c tout << n << I1 % I1 -c-c d << I' = I' -c-c tout << "(I' -c-c q -c-c 'I) ('I -cc d << 'I) c-c n -x endl;

6. Any point on the plane which is the bisector of the dihedral angle between two planes is equidistant from the sides of the angle.

This r u n iterated 4 times: 30 - 7 = 23,23 - 7 = 16, 16 - 7 = 9, a n d 9 - 7 = 2. So the quotient is 4, a n d the remainder is 2. Note that the following relationship must always be true for integer division: (quotient) (denominator) + (remainder) = numerator

Str2 = "1600 pENNSYLVANIA aVENUE nw, wASHINGTON, dc 20500" Str3 = "1600 Pennsylvania Avenue NW, Washington, DC 20500" First = 21 Second = 18

This program illustrates the use of the six arithmetic operators:

Extension of Locus Principles The following dual statements involve the locus of points in a plane and in space. Locus in a Plane 1. The locus of points at a given distance from a given point is a circle having the given point as center and the given distance as radius. Locus in Space 1. The locus of points at a given distance from a given point is a sphere having the given point as center and the given distance as radius.