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egoroff_w [7]
2 years ago
14

What is the common ratio of the geo netric sequence whose second and fourth terms are 6 and 54, respectively?

Mathematics
1 answer:
erastova [34]2 years ago
4 0
Try 2 as it is a factor of 6 and 54
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Does each function describe exponential growth or decay?
zhannawk [14.2K]

Hey Jinx :)

------------------------------------------------

Note that a decay has a rate between 0 and 1, and a growth as a rate of 1 or higher.

y = 1/2(1 + 0.03)^t → y = 1/2(1.03)^t (Rate is more then 0) GROWTH

y = 0.3(0.95)^t → (Rate is between 0 and 1) DECAY

y = ((1+0.03)^1/2)^2t → y = y = 1.014^2t (Rate is more then 0) GROWTH

y = 200(0.73)^t → (Rate is between 0 and 1) DECAY

y = 4(1/4)^t → (Rate is between 0 and 1) DECAY

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Hope This Helped! Good Luck!

4 0
3 years ago
Find the slope of the line through each pair of points.3) (3, -10), (14, -10)
mariarad [96]

Answer:

the second answer

Step-by-step explanation:

I took the test

4 0
2 years ago
Perform the indicated row operations, then write the new matrix.
Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]

Operations:

2R_1 + R_2 ->R_2

-3R_1 +R_3 ->R_3

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

The first operation:

2R_1 + R_2 ->R_2

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by 2

2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}0&5&7|1\end{array}\right]

The second operation:

-3R_1 +R_3 ->R_3

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by -3

-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

3 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=81-%28y-1%29%28y%5E2%2B27%29%20%5Cgeq%200%20%5C%5C%5C%5C%2081-%28y%5E3%2B27y-y%5E2-27%29%20%5C
Zarrin [17]
<span>First thing you'll need to know is that the value for this equation is actually an approximation 'and' it is imaginary, so, one method is via brute force method.

You let f(y) equals to that equation, then, find the values for f(y) using values from y=-5 to 5, you just substitute the values in you'll get -393,-296,-225,... till when y=3 is f(y)=-9; y=4 is f(y)=48, so there is a change in </span><span>signs when 'y' went from y=3 to y=4, the answer is between 3 and 4, you can work out a little bit deeper using 3.1, 3.2... You get the point. The value is close to 3.1818...

The other method is using Newton's method, it is similar to this but with a twist because it involves differentiation, so </span>y_{n+1}=y_n-\frac{f(y)}{f'(y)}<span> where 'n' is the number you approximate, like n=0,1,2... etc. f(y) would the equation, and f'(y) is the derivative of f(y), now what you'll need to do is substitute the 'n' values into 'y' to find the approximation.</span>
7 0
3 years ago
What is the minimum number of pieces of information you think you would need in order to prove that a given quadrilateral is a s
pashok25 [27]

Answer:

At least 4 numbers of information is needed

The four are highlighted below in the explanation section

Step-by-step explanation:

1. Given that the shape has four right angles and that all its sides are congruent then it's a quadrilateral

2. If two sides are consecutively congruent it is a quadrilateral shape

3. If a rhombus contains a right angle, then it’s a square shaped rhombus

4. If a quadrilateral is both a rectangle and a rhombus, then it’s a square

3 0
3 years ago
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