Bluebook SAT Test Digital SAT Hard Math Practice 4

Question 1

If 3^(x+1) = 9^(x−2), what is x?

Accepted Answer

5

Answer

9^(x+2) = 3^(2(x+2)) = 3^(2x+4); setting 2x−1 = 2x+4 still gives no solution. Correcting: 3^(x+1) = 9^(x−2) → x+1 = 2x−4 → x=5.

Question 2

A parabola with vertex (2, −3) passes through (5, 6). What is the leading coefficient a in f(x) = a(x − 2)² − 3?

Accepted Answer

1

Answer

Substituting (5, 6): 6 = a(5−2)² − 3 = 9a − 3, so 9a = 9 and a = 1.

Question 3

The function h(t) = −4.9t² + 14.7t + 2 models the height (in meters) of a ball after t seconds. What is the total time the ball is above a height of 2 meters?

Accepted Answer

3 seconds

Answer

Set h = 2: −4.9t² + 14.7t = 0 → t(−4.9t + 14.7) = 0 → t = 0 or t = 3; the ball is above 2 m for 3 − 0 = 3 seconds.

Question 4

If the inverse of f(x) = (3x + 2)/(x − 1) is f⁻¹(x), what is f⁻¹(5)?

Accepted Answer

7/2

Answer

Find x when f(x) = 5: (3x+2)/(x−1) = 5 → 3x + 2 = 5x − 5 → 7 = 2x → x = 7/2.

Question 5

A circle in standard form is (x − 3)² + (y + 2)² = 25. A line y = 4x − 2 intersects the circle. How many intersection points are there?

Accepted Answer

2

Answer

Substitute y = 4x − 2 into the circle: (x−3)² + (4x)² = 25 → 17x² − 6x + 9 − 25 = 0 → 17x² − 6x − 16 = 0; discriminant = 36 + 4·17·16 = 36 + 1088 = 1124 > 0 → two intersections.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Digital SAT Hard Math Practice 4