Solve a set of diverse math problems, including circles, rectangles, squares, right triangles, and profit calculations. Step-by-step solutions with detailed explanations are provided to help you understand and master each problem.
The circumference of a circle is equal to \( 72 \pi \). Find the radius of this circle.
The circumference of a circle is given by \[ C = 2 \pi r, \] where \( r \) is the radius. Substituting \( C = 72 \pi \) gives \[ 72 \pi = 2 \pi r. \] Solving for \( r \) gives \[ r = 36. \]
The length of a rectangular garden is 2 feet longer than 3 times its width. If the perimeter of the garden is 100 feet, find the width and the length.
Let \( L \) and \( W \) be the length and width of the garden. Then \[ L = 3 W + 2, \quad P = 2L + 2W. \] Substituting \( P = 100 \) and \( L = 3W + 2 \) gives \[ 100 = 2(3W + 2) + 2W \implies 100 = 8W + 4. \] Solving for \( W \) gives \[ W = 12, \quad L = 3(12) + 2 = 38. \]
A rectangular field has a length 10 feet more than its width. If the area of the field is 264, what are the dimensions?
Let \( x \) and \( y \) be the length and width. Then \[ x = y + 10, \quad A = xy = 264. \] Substituting gives \[ 264 = (y + 10)y \implies y^2 + 10y - 264 = 0. \] Solving yields \[ y = 12 \quad (\text{reject negative solution}), \quad x = y + 10 = 22. \]
A company's revenue is \( x^2 + 100x \) dollars and costs are \( 240x + 500 \) dollars. How many units must be sold to make a profit of 10,000 dollars?
Profit formula: \[ P = \text{Revenue} - \text{Cost} = (x^2 + 100x) - (240x + 500) = x^2 - 140x - 500. \] Set \( P = 10000 \): \[ 10000 = x^2 - 140x - 500 \implies x^2 - 140x - 10500 = 0. \] Solving gives \[ x = 194 \quad (\text{positive solution only}). \]
A square has a side 5 cm shorter than a second square. The area of the larger square is four times the area of the smaller square. Find the side of each square.
Let \( x \) and \( y \) be the sides of the smaller and larger squares: \[ x = y - 5, \quad y^2 = 4x^2 = 4(y - 5)^2. \] Simplify: \[ y^2 - 4(y - 5)^2 = 0 \implies (y - 10)(y - 10/3) = 0. \] Accept positive solution: \[ y = 10, \quad x = y - 5 = 5. \]
Find two numbers whose sum is 26 and product is 165.
Let the numbers be \( a \) and \( b \): \[ a + b = 26, \quad ab = 165 \implies b = 26 - a. \] Substitute: \[ a(26 - a) = 165 \implies a^2 - 26a + 165 = 0. \] Solve: \[ a = 11, b = 15 \quad \text{or} \quad a = 15, b = 11. \]
The area of a rectangle is 15 and the perimeter is 16. Find its dimensions.
Let \( x \) and \( y \) be the sides: \[ xy = 15, \quad 2x + 2y = 16 \implies x + y = 8. \] Solve for \( x = 8 - y \), substitute: \[ (8 - y)y = 15 \implies y^2 - 8y + 15 = 0. \] Solutions: \[ y = 3, x = 5 \quad \text{or} \quad y = 5, x = 3. \]
The sum of two numbers is 20. The larger number is four less than twice the smaller number. Find the numbers.
Let \( x \) and \( y \) be the larger and smaller numbers: \[ x = 2y - 4, \quad x + y = 20. \] Substitute: \[ 2y - 4 + y = 20 \implies 3y = 24 \implies y = 8, \quad x = 12. \]
The hypotenuse of a right triangle is 2 cm more than the longer side. The shorter side is 7 cm less than the longer side. Find the hypotenuse.
Let \( h \) be the hypotenuse, \( a \) the longer side, \( b \) the shorter side: \[ h = a + 2, \quad b = a - 7. \] Pythagoras: \[ h^2 = a^2 + b^2 \implies (a + 2)^2 = a^2 + (a - 7)^2. \] Simplify: \[ a^2 + 4a + 4 = a^2 + a^2 - 14a + 49 \implies a^2 - 18a + 45 = 0. \] Solve for \( a \) and then \( h = a + 2 = 17 \) cm.