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2 years ago

Which of these tables represent a nonlinear function?

Mathematics

2 answers:

Nastasia [14]2 years ago

6 0

The third one. As the x increases by 1, the increase in y is not constant

ddd [48]2 years ago

3 0

Answer:

(3)

Step-by-step explanation:

Analise the slope m = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

(1). (19-20)/(18-17) = - 1 ; (18-19)/(19-18) = - 1 ; (17-18)/(20-19) = - 1 ⇒ table represents a linear function.

(2). (-17+16)/(18-17) = - 1 ; (-18+17)/(19-18) = - 1 ⇒ table represents a linear function.

(3). (17-16)/(18-17) = 1 ; (19-17)/(19-18) = 2 ⇒ table represents a nonlinear function.

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What is the angular velocity of a 6–foot pendulum that takes 3 seconds to complete an arc of 14.13 feet? Use 3.14 for π

iren2701 [21]

1) Calculate the whole arc (complete circumference):

Circumference = 2π*radius

radius = 6 foot

=> Circumference = 2 * 3.14 * 6 foot = 37.68

2) Calculate the time to make a whole round, by making a proportion with the two ratios:

ratio 1 = x / 37.68 =

ratio 2 = 3 / 14.13

ratio 1 = ratio 2 =>

   x 3s
---------- =   -----------
37.68 ft    14.13 ft

=> x = 3 * 37.68 / 14.13 = 8 s

3) Calcuate the angular velocity as the 2π rad (which is the total angle of a circle) divided by the time.

So the angular velocity is 2π rad / 8 s = 2*3.14 foot / 8 s = 0.785 rad / s

Answer: 0.785 rad / s

3 0

3 years ago

Find x? In 3x - In(x - 4) = ln(2x - 1) +ln3

earnstyle [38]

Answer:

$x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

Step-by-step explanation:

Because $3\, x$ is found in the input to a logarithm function in the original equation, it must be true that $3\, x > 0$ . Therefore, $x > 0$ .

Similarly, because $(x - 4)$ and $(2\, x - 1)$ are two other inputs to the logarithm function in the original equation, they should also be positive. Therefore, $x > 4$ .

Let $a$ and $b$ represent two positive numbers (that is: $a > 0$ and $b > 0$ .) The following are two properties of logarithm:

$\displaystyle \ln (a) + \ln(b) = \ln\left(a \cdot b\right)$ .

$\displaystyle \ln (a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ .

Apply these two properties to rewrite the original equation.

Left-hand side of this equation:

$\begin{aligned}&\ln(3\, x) - \ln(x - 4)= \ln\left(\frac{3\, x}{x -4}\right)\end{aligned}$

Right-hand side of this equation:

$\ln(2\, x- 1) + \ln(3) = \ln\left(3 \left(2\, x - 1\right)\right)$ .

Equate these two expressions:

$\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ .

The natural logarithm function $\ln$ is one-to-one for all positive inputs. Therefore, for the equality $\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ to hold, the two inputs to the logarithm function have to be equal and positive. That is:

$\displaystyle \frac{3\ x}{x - 4} = 3\, (2\, x - 1)$ .

Simplify and solve this equation for $x$ :

$x^2 - 5\, x + 2 = 0$ .

There are two real (but not rational) solutions to this quadratic equation: $\displaystyle \frac{5 + \sqrt{17}}{2}$ and $\displaystyle \frac{5 - \sqrt{17}}{2}$ .

However, the second solution, $\displaystyle \frac{5 - \sqrt{17}}{2}$ , is not suitable. The reason is that if $x = \displaystyle \frac{5 - \sqrt{17}}{2}$ , then $(x - 4)$ , one of the inputs to the logarithm function in the original equation, would be smaller than zero. That is not acceptable because the inputs to logarithm functions should be greater than zero.