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3 years ago

13

Find the missing terms of the following geometric sequence. Assume all terms are positive. (Hint: The geometric mean of the fi

rst and fifth terms is the third term.)
1.5,_,_,_,121.5....

1 answer:

alexandr1967 [171]3 years ago

7 0

1.5,4.6,13.5,40.5,121.5

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What is the distance from the point (0,-4) to point (4,4)?

dlinn [17]

Answer: Distance is 8.9

Step-by-step explanation:

$d = √((x2-x1)2 + (y2-y1)2)$

$(x2-x1) = (4 - 0) = 4$

$(y2-y1) = (4 -4) = 8$

Add it all together

$(4)2 + (8)2 = 16 + 64 = 80$

Square 80

$\sqrt{80} =4\sqrt{5} \\=8.9443$

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4 years ago

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How many possible outcomes exist when two fair coins are flipped and a three-section spinner is spun?

kodGreya [7K]

Each coin can have the outcomes heads (H) or tails (T).
The spinner can have the outcomes 1, 2, 3.
The total number of outcomes is the product of the individual numbers of outcomes.
2 * 2 * 3 = 12

Here's a way of thinking about it:

The coins can have these four outcomes:
HH
HT
TH
TT
For each of the 4 outcomes of the two coins, there are 3 different outcomes of the spinner, so the total number of possible outcomes is 4 * 3 = 12.

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3 years ago

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A florist has 8 roses and 24 daffodils.she wants to make as nany identical bouquets as possible, with the same combination of ro

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They should have 3 daffodils and 1 rose in each bouquet and that will equal 8 bouquets.

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In the bleachers at the basketball game,1/4 of the fans are adult men, and 5/12 are adult women. What fraction of the fans are a

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1/3 are children and 2/3 are adults

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3 years ago

Use a linear system to write u = (12, 19, 31) as a linear combination of U1 = (1,1,2), u2 = (2,3,5), and uz = (3,5,8). Is w = (1

KiRa [710]

Answer:

Let $t\in {\mathbb R}$ ,

$\vec{u} = (t-2)\vec{u}_1 + (-2t +7)\vec{u}_2 + t\cdot \vec{u}_3$ .

$\vec{w}$ is also a linear combination of $\vec{u}_1$ , $\vec{u}_2$ , $\vec{u}_3$ .

Step-by-step explanation:

<h3>1.</h3>

Write a linear system for $\vec{u} = x_1\cdot\vec{u}_1 + x_2\cdot \vec{u}_2 + x_3\cdot \vec{u}_3$ , with one equation for each component. The augmented matrix for the first linear system will be:

$\displaystyle \left[\begin{array}{ccc|c}1 & 2 & 3 & 12\\1 & 3 & 5 & 19\\2 & 5 & 8 & 31\end{array}\right]$ .

Transform this matrix to its reduced row-echelon form using Gaussian Elimination. Solve for each variable.

$\begin{aligned} &\left[\begin{array}{ccc|c}1 & 2 & 3 & 12\\1 & 3 & 5 & 19\\2 & 5 & 8 & 31\end{array}\right]\\ &\sim \left[\begin{array}{ccc|c}1 & 2 & 3 & 12\\0 & 1 & 2 & 7\\0 & 1 & 2 & 7\end{array}\right]\\&\sim \left[\begin{array}{ccc|c}1 & 2 & 3 & 12\\0 & 1 & 2 & 7\\0 & 0 & 0 & 0\end{array}\right]\\&\sim\left[\begin{array}{ccc|c}1 & 0 & -1 & -2\\0 & 1 & 2 & 7\\0 & 0 & 0 & 0\end{array}\right]\\&\left\{\begin{array}{l}x_1 = t-2\\x_2=- 2t+7\\x_3=t\end{array}\right.\end{aligned}$ .

Therefore,

$\vec{u} = (t-2)\vec{u}_1 + (-2t +7)\vec{u}_2 + t\cdot \vec{u}_3$ .

<h3>2.</h3>

Set up a similar augmented matrix for $\vec{w} = x_1\cdot\vec{u}_1 + x_2\cdot \vec{u}_2 + x_3\cdot \vec{u}_3$ :

$\left[\begin{array}{ccc|c}1 & 2 & 3 & 1\\1 & 3 & 5 & 0\\2 & 5 & 8 & 1\end{array}\right]$ .

The second part of this question isn't concerned about the exact value of $x_1$ , $x_2$ , or $x_3$ . Therefore, before proceeding with Gaussian Elimination, start by checking the determinant of the coefficient matrix. If this determinant is nonzero, $\vec{w}$ will always be a unique linear combination of $\vec{u}_1$ , $\vec{u}_2$ , $\vec{u}_3$ now matter what value it takes.

In this case (also as seen in the first part of this question), the determinant of the coefficient matrix for $\vec{u}_1$ , $\vec{u}_2$ , and $\vec{u}_3$ is zero. Determining whether the linear combination is possible will require elimination.

$\begin{aligned} &\left[\begin{array}{ccc|c}1 & 2 & 3 & 1\\1 & 3 & 5 & 0\\2 & 5 & 8 & 1\end{array}\right]\\ &\sim \left[\begin{array}{ccc|c}1 & 2 & 3 & 1\\0 & 1 & 2 & -1\\0 & 1 & 2 & -1\end{array}\right]\\&\sim \left[\begin{array}{ccc|c}1 & 2 & 3 & 1\\0 & 1 & 2 & -1\\0 & 0 & 0 & 0\end{array}\right]\\&\sim\left[\begin{array}{ccc|c}1 & 0 & -1 & 3\\0 & 1 & 2 & -1\\0 & 0 & 0 & 0\end{array}\right]\\&\left\{\begin{array}{l}x_1 = t+3\\x_2=- 2t-1\\x_3=t\end{array}\right.\end{aligned}$ .

Similar to the first part of this question, this linear system is consistent. $\vec{w} = (t+3)\vec{u}_1 + (-2t -1)\vec{u}_2 + t\cdot \vec{u}_3$ . $\vec{w}$ is indeed a linear combination of $\vec{u}_1$ , $\vec{u}_2$ , $\vec{u}_3$ .

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3 years ago