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marshall27 [118]
3 years ago
11

Determine the intervals on which the function is increasing, decreasing, and constant. An absolute value function is shown facin

g down with a vertex of -1,0.

Mathematics
1 answer:
gavmur [86]3 years ago
5 0

Answer:

Increasing on: x\:

Decreasing on: x\:>\:-1

Constant at : x=-1

Step-by-step explanation:

The given absolute value function is f(x)=|x+1|.

A function is said to be increasing if for all x_1\:>\:x_0,  f(x_1)\:>\:f(x_0)

From the graph, we can observe that the graph has a positive slope for all x-values less than -1. This implies that the interval of increase is x\: or (-\infty,-1)

We can also observe that, the slope of this function is negative on the interval:x\:>\:-1 or (-1,\infty).

At x=-1, the function is neither increasing nor decreasing. We say the function is constant at x=-1

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1+1<br> NEED HELP LOL<br> urgent<br> help
navik [9.2K]

Answer:

<h3>2</h3>

Step-by-step explanation:

\mathrm{Add\:the\:numbers:}\:1+1=2\\\\

4 0
3 years ago
Read 2 more answers
Help with this one please 20 points brain lest and a thanks on your page honestly just giving points away
Kitty [74]
The answer is C.

It has a negative slope (-40) and a y-intercept at 200.
3 0
4 years ago
Find the general solution of {y}''-3y=8e^{3t}+4sin(t) .
babunello [35]

Answer:

y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint

Step-by-step explanation:

We are given that linear differential equation

y''-3y=8e^{3t}+4 sint

Auxillary equation

D^2-3=0

D=\pm \sqrt3

C.F=C_1e^{\sqrt3t}+C_2e^{-\sqrt3}

P.I=\frac{8e^{3t}+4sin t}{D^2-3}

P.I=\frac{8e^{3t}}{9-3}+4\frac{sint }{-1-3}

P.I=e^{ax}{\phi (D+a)} and P.I=\frac{sinax}{(\phi D)}where D square is replace by - a square

P.I=-\frac{4}{3}e^{3t}- sint

Hence, the general solution

G.S=C.F+P.I

y(x)=C_1e^{\sqrt3t}+C_2e^{-\sqrt3t}-\frac{4}{3}e^{3t}-sint

3 0
3 years ago
What is the slope of a linear that is perpendicular to the line y=-1/2x+5?​
Vlada [557]

Step-by-step explanation:

Equation of given line is:

y =  -  \frac{1}{2} x + 5 \\  \\ equating \: it \: with \: y =m_1x + c  \\ we \: find \:  \\ m_1 = -  \frac{1}{2} \\ let \: m_2 \: be \: the \: slope \: of \: required \: line \\  \\  \therefore \: m_1 \times m_2 =  - 1( \because \: lines \: are \:  \perp) \\  \\ \therefore \: -  \frac{1}{2} \times m_2 =  - 1\\  \\ \therefore \: m_2 = 2 \\

Thus, the slope of required line is 2.

6 0
3 years ago
EMERGENCY 11??? DUE TOMORROW
poizon [28]

First, we plug in the numbers.

$40 = P(4%)(2)

4% = 0.04

40 = P * 0.04 * 2

Divide each side by 2

20 = P * 0.04

Divide by 0.04

P = 500

Lets check it!

500 * 0.04 * 2 = 40

500 * 0.04 = 20

20 * 2 = 40

40 = 40

8 0
3 years ago
Read 2 more answers
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