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photoshop1234 [79]
3 years ago
8

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?

Mathematics
1 answer:
Greeley [361]3 years ago
5 0

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

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Step-by-step explanation:

All we can do with this equation is factor it.

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When we look at the coefficients, there is a common factor of 4 with 16 and 4. The p's are also common factors, and we can take out a common factor of x^3. We can combine these common factors and take them out of the equation at the same time.

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Answer:

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Step-by-step explanation:

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