During a 5 -second time interval, the average acceleration a , in meters per second squared, of an object with an initial velocity of 12 meters per second is defined by the equation a = (vf - 12)/5 , where vf is the final velocity of the object in meters per second. If the equation is rewritten in the form vf = xa + y , where x and y are constants, what is the value of x ?

The correct answer is 5. The correct answer is 5 . The given equation can be rewritten in the form vf = xa + y , like so: a = (vf - 12)/5 vf - 12 = 5a vf = 5a + 12 It follows that the value of x is 5 and the value of y is 12 .

Note: The listed acceleration values let the regression fit vf=wa+q, so w is the requested coefficient of acceleration.

SAT Question f2f3fa00: Answer, Explanation & Desmos Solution

During a $5$ -second time interval, the average acceleration $a$ , in meters per second squared, of an object with an initial velocity of $12$ meters per second is defined by the equation $a = \frac{v _{f} - 12}{5}$ , where $v_{f}$ is the final velocity of the object in meters per second. If the equation is rewritten in the form $v_{f} = x a + y$ , where $x$ and $y$ are constants, what is the value of $x$ ?

148

Difficulty:Hard

SAT Question f2f3fa00: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Explanation

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