What is the rate of change for the linear relationship modeled in the table? (4 points)

Mathematics

2 answers:

Karolina [17]3 years ago

8 0

Answer:1/2

Step-by-step explanation:

Because the other guy said so and they seem smart so...yeah I would pick that one.

lozanna [386]3 years ago

5 0

Answer:

-1/2

Step-by-step explanation:

In a linear relationship, the rate of change of one variable with respect to the other is constant. When we talk about change, we're looking for a difference of values.

If we look at the first and second rows, the change in x is 1 - (-1) = 2, while the change in y is 9 - 10 = -1. Usually we refer to these changes as Δx and Δy (read like "delta-x" and "delta-y"), and the rate of change is the number we get by dividing one of these by the other.

The rate of change we're used to seeing, sometimes called the slope, is Δy/Δx. So, using the values we've already found:

$\Delta{x}=2\\\Delta{y}=-1\\\Delta{y}/\Delta{x}=-1/2$

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H(x)=1/8x^3-x^2 Over which interval does h have a positive average rate of change?

lara31 [8.8K]

Answer:

$(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}$

Step-by-step explanation:

$h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \ respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x$

For positive rate of change $h'(x)>0$

$\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x\frac{16}{3} \\\\$