II. Let f(x) = 9 – x , g(x) = x2 + 4, and h(x) = x – 2. Compute the following: 7. g(f(12))

Mathematics

1 answer:

Alex17521 [72]2 years ago

4 0

G(f(12))
f(12) = 9 - 12 = -3
g(-3) = (-3)^2 + 4 = 9 + 4 = 13

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Enter the correct answer in the box. What are the solutions of this quadratic equation? X^2+164=16x Substitute the values of a a

saw5 [17]

Answer:

x = 8 - 10i or 8 + 10i

Step-by-step explanation:

Our equation is given as:

x² + 164 = 16x

x² - 16x + 164 = 0

We solve using Almighty formula

x = -b ± √b² - 4ac/2a

Where : ax² + bx + c

For our Equation

a = 1, b = -16, c = 164

Hence,

x = -(-16) ± √-16² - 4 × 1 × 164/2 × 1

x = 16 ± √256 - 656/2

x = 16 ± √-400/2

Finding the roots

x = 16/2 ± √-400/2

x = 8 ± 20i/2

x = 8 ± 10i

Therefore, the value of x

= 8 - 10i or 8 + 10i

4 0

2 years ago

A: 2.86<br> B: 2.75<br> C: 2.95<br> D: 1.17

zhuklara [117]

No doubt your text advises how to arrive at a best-fit exponential model. The two graphing calculators I used gave different answers. The first one modeled the function as
y = 398.494(0.50526)^x
For x = 7, this model gives y = 3.35

The second calculator, a TI-84 work-alike, modeled the function as
y = 422.110(0.48896)^x
For x=7, this model gives y = 2.82

Using the first and last data points only, the model is
y = 200(0.06)^((x-1)/4)
For x=7, this model gives y = 2.94

Using yet another calculator to do linear regression on the log of y, the function is modeled by
y = 10^(2.62543-0.31073x)
which is another way to write the function produced by the TI-84 work-alike.

So, the technology I used gives a couple of reasons to choose
A. 2.86
and a couple of reasons to choose
C. 2.95

Of the models shown here, the ones with the lowest mean absolute deviation and the least squared error were the ones that produced the highest values of y for x=7, suggesting 2.95 is the best choice.

It would be advisable to use the procedure described in your text.