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skelet666 [1.2K]

3 years ago

9

Identify one linear equation and one nonlinear equation, and state why it is linear or nonlinear. Thanks

1 answer:

alukav5142 [94]3 years ago

6 0

Equation (2) y = 2x - 5 is linear equation.

Equation with the highest power of x = 1 is a linear equation.

It is linear because it forms a straight line.

Equation (3) y = x² + 6 is not a linear equation.

Equation with the highest power of x = 2 is a curve, specifically a parabola curve.

It is not a linear equation because it is a curve.

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Which pair of expressions below are equivalent? a. −4−8 and 4+8 b. −4−8 and −4+8 c. −4−8 and −4−(−8) d. −4−8 and −4+(−8)

omeli [17]

Answer:

d. −4−8 and −4+(−8)

Step-by-step explanation:

+ - is equivalent to -

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Nevermind give the points to the another guy

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The slope f '(x) at each point (x, y) on a curve. y = f (x) is given along with a particular point (a, b) on the curve. Use this

Advocard [28]

Answer:

Step-by-step explanation:

Integrating each term with respect to x, we get:

x^3 x^2

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3 2

We are told that if x = 0, f(x) = -7, and so C must equal - 7.

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3 years ago

Please help. linear equations.

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3 years ago

Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a

Bond [772]

$f(x)=3\sin x+3\cos x\implies f'(x)=3\cos x-3\sin x$

$f$ has critical points where $f'=0$ :

$3\cos x-3\sin x=0\implies\cos x=\sin x\implies\tan x=1\implies x=\pm\dfrac\pi4+2n\pi$

where $n$ is any integer. We get solutions in the interval $\left[0,\frac\pi3\right]$ for $n=0$ , for which $x=\frac\pi4$ .

At this critical point, we have $f\left(\frac\pi4\right)=3\sqrt2\approx4.243$ .

At the endpoints of the given interval, we have $f(0)=3$ and $f\left(\frac\pi3\right)=\frac{3+3\sqrt3}2\approx4.098$ .

So we have the extreme values

$\max\limits_{x\in\left[0,\frac\pi3\right]}f=3\sqrt2$

$\min\limits_{x\in\left[0,\frac\pi3\right]}f=3$

8 0

3 years ago