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Firdavs [7]
3 years ago
5

Solve for x

1" title=" \frac{x}{a} = \frac{x}{b} " alt=" \frac{x}{a} = \frac{x}{b} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
lesya [120]3 years ago
6 0

Domain:\ a\neq0\ \wedge\ b\neq0

\dfrac{x}{a}=\dfrac{x}{b}\qquad|\text{cross multiply}\\\\bx=ax\qquad|\text{subtract ax from both sides}\\\\bx-ax=0\\\\x(b-a)=0\qquad|\text{divide both sides by }\ b-a\neq0\\\\x=0

\dfrac{x}{a}=\dfrac{x}{b}\\\\if\ a\neq b\ then\ x=0\\\\if\ a=b\ then\ x\in\mathbb{R}

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Lesson: 1.08Given this function: f(x) = 4 cos(TTX) + 1Find the following and be sure to show work for period, maximum, and minim
Ber [7]

The given function is

f(x)=4\cos \text{(}\pi x)+1

The general form of the cosine function is

y=a\cos (bx+c)+d

a is the amplitude

2pi/b is the period

c is the phase shift

d is the vertical shift

By comparing the two functions

a = 4

b = pi

c = 0

d = 1

Then its period is

\begin{gathered} \text{Period}=\frac{2\pi}{\pi} \\ \text{Period}=2 \end{gathered}

The equation of the midline is

y_{ml}=\frac{y_{\max }+y_{\min }}{2}

Since the maximum is at the greatest value of cos, which is 1, then

\begin{gathered} y_{\max }=4(1)+1 \\ y_{\max }=5 \end{gathered}

Since the minimum is at the smallest value of cos, which is -1, then

\begin{gathered} y_{\min }=4(-1)+1 \\ y_{\min }=-4+1 \\ y_{\min }=-3 \end{gathered}

Then substitute them in the equation of the midline

\begin{gathered} y_{ml}=\frac{5+(-3)}{2} \\ y_{ml}=\frac{2}{2} \\ y_{ml}=1 \end{gathered}

The answers are:

Period = 2

Equation of the midline is y = 1

Maximum = 5

Minimum = -3

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