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3 years ago

Can somebody simply explain a unlinear functions???

Mathematics

2 answers:

wlad13 [49]3 years ago

5 0

Answer:

The graph of a linear function is a line. ... Linear functions have a constant slope, so nonlinear functions have a slope that varies between points. Algebraically, linear functions are polynomials with highest exponent equal to 1 or of the form y = c where c is constant. Nonlinear functions are all other functions.

nonlinear functions are functions that are not linear functions. Therefore, they have the opposite properties of a linear function. The graph of a linear function is a line. Thus, the graph of a nonlinear function is not a line.

Step-by-step explanation:

sertanlavr [38]3 years ago

3 0

Answer Algebraically, linear functions are polynomials with highest exponent equal to 1 or of the form y = c where c is constant. Nonlinear functions are all other functions. An example of a nonlinear function is y = x^2. This is nonlinear because, although it is a polynomial, its highest exponent is 2, not 1.

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3 years ago

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4.. Please help. Is DE←→ perpendicular to FG←→? Why or why not?

olya-2409 [2.1K]

Given:

DE contains the points D(1, -2) and E(3, 4).

FG contains the points F(-1, 2) and G(4, 0).

To find:

Is DE perpendicular to FG.

Solution:

Slope of DE:

$$m=\frac{y_2-y_1}{x_2-x_1}$

Here $x_1=1, y_1=-2, x_2=3, y_2=4$

$$m=\frac{4-(-2)}{3-1}$

$$m=\frac{6}{2}$

m = 3

Slope of FG:

$$m=\frac{y_2-y_1}{x_2-x_1}$

Here $x_1=-1, y_1=2, x_2=4, y_2=0$

$$m=\frac{0-2}{4-(-1)}$

$$m=\frac{-2}{4+1}$

$$m=\frac{-2}{5}$

Two lines are perpendicular if product their slopes are -1.

Slope of DE × Slope of FG

$$=3\times \frac{-2}{5}$

$$= \frac{-6}{5}$

≠ -1

The solution is no, because the product of the slopes is not -1.

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3 years ago

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PLEASE HELP NOW IM FAILING MATH

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X = 89

A triangle always equals 180°. Because two out of three angles are given, the two known angles (42° and 49°) can be added together which is 91°. The 91° can be subtracted from 180° to find the unknown angle x.

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3 years ago

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A thermometer is taken from an inside room to the outside, where the air temperature is 20° f. after 1 minute the thermometer re

fomenos

The change in the temperature is proportional to the difference in temperatures. Let's call the initial temperature of the room and the thermometer $E_0$ and the current temperature $E(t)$ where $t$ is the elapsed time in minutes. The outside temperature is 20 F.

Newton's Law of Cooling:

$E(t) - 20 = (E_0 - 20) e^{- k t}$

We're given $E(1)=70 \textrm{ and } E(5)=45$

$70-20 = (E_0 - 20) e^{-k}$

$45-20 =(E_0 - 20) e^{-5k}$

Dividing,

$\dfrac{50}{25}=e^{4k}$

$k = \dfrac 1 4 \ln 2$

Now,

$E(t) - 20 = (E_0 - 20) e^{- k t}$

$E_0 = 20 + (E(t) - 20)e^{k t}$

$E_0 = 20 + (E(1) - 20)e^{k}$

$E_0 = 20 + (70 - 20)e^{\ln 2/4}$

$E_0 = 20 + 50 \cdot 2^{1/4} \approx 79.46 \textrm{ degrees F}$

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4 years ago

Can somebody simply explain a unlinear functions???​

Can somebody simply explain a unlinear functions???