If F(a, b, c, d) = a^b + c \times d, what is the value of x such that F(2, x, 4, 11) = 300?
1 answer:
Answer:
x = 8
Step-by-step explanation:
A graphing calculator can show you the answer easily. It works well to define a function whose x-intercept is the solution. We can do that by subtracting 300 from the given equation so we have ...
F(2, x, 4, 11) -300 = 0
The solution is x = 8.
__
We can solve this algebraically:
F(2, x, 4, 11) -300 = 0
2^x +4·11 -300 = 0 . . . . use the function definition
2^x -256 = 0 . . . . . . simplify
2^x = 2^8 . . . . . add 256
x = 8 . . . . . . . . . match exponents of the same base
You might be interested in
Answer:
7 units
Step-by-step explanation:
The formula for a circle is: , where (x_1, y_1) is the center of the circle and r is the radius.
Right now, we see that r^2 = 49, so to find r, we square root 49 and get 7.
Thus, the answer is 7 units.
Hope this helps!