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2 years ago

Write the formula for an exponential function with initial value 300 and decay factor 0.73

Mathematics

1 answer:

postnew [5]2 years ago

8 0

Answer: 300*0.73^x

Step-by-step explanation:

300 is the initial value and every time x goes up by 1, the value get multiplied by 0.73

when x=3

then the value would be 300*0.73*0.73*0.73 or 300*0.73^3

then for x amount of times the equation would be 300*(0.73 x times ) or 300*0.73^x

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Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

——————————

Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

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Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

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2 years ago

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Afina-wow [57]

Find a common denominator which is 12

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then 5*2=10 6*2=12 so 2 10/12

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2 years ago

Read 2 more answers

Help!!!!!!!!!!!!!!!!!!!!

lapo4ka [179]

Well... the segment FG = 4x + 3, the segment GH = 7x - 12, now, the

segment FH = FG + GH, those two fellows added up, and we know that FG and GH are both equal halves because G is the midpoint.

thus, FG = GH
or

$\bf 4x+3=7x-12\implies 3+12=7x-4x\implies 15=3x
\\\\\\
\cfrac{15}{3}=x\implies \boxed{5=x}\\\\
-------------------------------\\\\
thus\qquad 
\begin{cases}
\overline{FG}=4x+3\\
4(5)+3\\
23\\
------\\
\overline{GH}=7x-12\\
7(5)-12\\
23
\end{cases}\impliedby equal\ halves
\\\\\\
\overline{FH}=23+23\implies \overline{FH}=46\implies 6y-2=46
\\\\\\
6y=46+2\implies y=\cfrac{48}{6}$

and fairly sure, you know how much that is.

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2 years ago