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.
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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
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Answer: B. H(t) = -6.9(t - 2.3)² + 112
<u>Step-by-step explanation:</u>
The vertex form of a quadratic equation is: y = a(x - h)² + k
where (h, k) is the vertex ⇒ k is the maximum height
the distance traveled is when y = 0 ⇒
Given: H(t) = -7.1(t - 2.3)² + 98
maximum height (k) = 98 feet
distance traveled (x) = = 6.02 seconds
A) H(t) = -7.5(t - 2.2)² + 112
maximum height (k) = 112 feet
distance traveled (x) = = 6.06 seconds
B) H(t) = -6.9(t - 2.3)² + 112
maximum height (k) = 112 feet
distance traveled (x) = = 6.33 seconds
C) H(t) = -6.9(t - 2.4)² + 95
maximum height (k) = 95 feet
This has a lower height than the given equation.
D) H(t) = -7.5(t - 2.3)² + 95
maximum height (k) = 95 feet
This has a lower height than the given equation.
Both options A and B travel higher and stay in the air longer than last year's winner, however option B stays in the air longer than option A.