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3 years ago

Given the recursive formula for a geometric sequence find the common ratio, the first four terms, and the explicit formula.

1 answer:

4 0

Since a(n)=-3(a(n-1)) the common ratio is -3, and a(1)=-3 so

a(n)=-3(-3^(n-1)) so the first four terms are:

-3, 9, -27, 81

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Last month, a dwarf lemon tree grew half as much as a semi-dwarf lemon tree. A full-size lemon tree grew three times as much as

dimulka [17.4K]

Answer: the dwarf tree grew by 3 inches.

the semi dwarf tree grew by 6 inches.

the full size tree grew by 18 inches.

Step-by-step explanation:

Let x represent how much the semi-dwarf lemon tree grew.

Last month, a dwarf lemon tree grew half as much as a semi-dwarf lemon tree. This means that the amount by which the dwarf lemon tree grew is expressed as x/2

A full-size lemon tree grew three times as much as the semi-dwarf lemon. This means that the amount by which the full-size lemon tree grew is expressed as 3x

Together, the three trees grew 27 inches. This means that

x/2 + x + 3x = 27

Cross multiplying by 2, it becomes

x + 2x + 6x = 54

9x = 54

x = 54/9

x = 6 inches

The dwarf tree grew by 6/2 = 3 inches.

The full-size lemon tree grew by 3 × 6 = 18 inches

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3 years ago

What is -3x+7y=13 in slope intercept form

Gekata [30.6K]

Answer : y = 3/7x + 13/7

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2 years ago

What goes up and down without moving?<br> Whoever answers first gets brainliest.

pav-90 [236]

Answer: temperature and weight

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2 years ago

Is Amy’s answer accurate

Rom4ik [11]

Answer:

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3 years ago

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

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Solve the trigonometric equation:

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

Make a substitution:

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

and the equation becomes

$\mathsf{2t^2-t-1=0}$

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. =)

3 0

3 years ago