To solve this problem, it might be helpful to make a list of the powers of 2: 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, and so on. Select consecutive values from this list and multiply them, starting with the values that make the most sense. For example, you know that 2 × 4 × 8 will be less than 512, so try 4 × 8 × 16, which does equal 512. Now, add those same terms: 4 + 8 + 16 = 28.

To solve this problem, use the formula for the area of a trapezoid,¹⁄₂h(b₁+b₂), where h is the altitude (height) and b₁ and b₂ are the bases. In this case, the altitude (height) is 3 and one of the bases is 5. Because the area is known to be 24, the other base can be determined by using the equation 24 =(¹⁄₂)(3)(b + 5). Dividing both sides of the equation by ³⁄₂ (or multiplying both sides of the equation by ²⁄₃) yields 16 = b + 5, or b = 11.

If the function f satisfies the equation f (x+y) = f (x)+f (y) for every pair of real numbers x and y, then f is a linear function. For some unknown linear function f , the value of f (1) can be any real number (think of all the possible lines that can be drawn).

To solve this problem, recall that circumference is 2πr, where r is the radius. If a circle has a circumference of (⁴⁄₃)π, then (⁴⁄₃)π = 2πr. Therefore, 2r = ⁴⁄₃, or r = ²⁄₃.

If the measure of angle DCA is 60°, then the measure of angle BCA is 60° as well. Because the angles in a triangle have the sum of 180°, angle BAC is 180 − 90 − 60, or 30°. Because the segment AD bisects angle BAC, angle BAD is equivalent to angle DAC, which is \(\frac{30^{\circ}}{2}\) , or 15°. Further, since the triangle ABD is a right triangle, the measure of angle ADB = 180 − 90 − 15, or 75°.

To solve this problem, recall that a quadratic equation has exactly one real solution when the solution is a “double zero.” Such an equation would have the factored form of (y + b)² = 0, where b is a real number. When you expand the equation, it becomes (y + b)² is y² + 2by + b². The equation y² + 2ky + 4k = 0 has a similar form; however, in order for the term 4k to represent b², k must equal 4.

Enclosing a square region with 120 feet of fence would make each side of the square \(\frac{120}{4}\) , or 30 feet. Therefore, the number of square feet enclosed is the area of the region, which would be 30², or 900 square feet.

The tangent of an angle is the ratio of the side opposite the angle to the side adjacent to the angle. The side opposite angle x is 15 and the side adjacent is 8, making tan x = \(\frac{15}{8}\).

To solve this problem, you must recall that a circle with center (h, k) and radius r has the equation (x − h)² + (y − k)² = r². A circle that has center (−4, 5) and radius 5 units would therefore have an equation (x + 4)² + (y − 5)² = 25.