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Artist 52 [7]
3 years ago
15

Eileen has a greenhouse and is growing orchids. The table shows the average number of orchids that bloomed over a period of four

months:
Month 1 2 3 4
Orchids 11 12.4 13.8 15.2


Did the number of orchids increase linearly or exponentially?
Linearly, because the table shows that the orchids increased by the same amount each month
Linearly, because the table shows a constant percentage increase in orchids each month
Exponentially, because the table shows that the orchids increased by the same amount each month
Exponentially, because the table shows a constant percentage increase in orchids each month
Mathematics
2 answers:
LenaWriter [7]3 years ago
8 0

Answer:

Linearly, because the table shows a constant percentage increase in orchids each month

Step-by-step explanation:

I took a test and got the same question

Anuta_ua [19.1K]3 years ago
3 0

Answer:

B

Step-by-step explanation:

I believe this is the answer because each orchids is increasing by 1.4 each month. Hopefully I'm right on this.

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Determine whether the given system has a unique solution, no solution, or infinitely many solutions.
victus00 [196]

Answer:

The system has no solution.

Step-by-step explanation:

To find the solution to this system of linear equations

\left\begin{array}{ccccc}-3x_1&+x_2&-2x_3&=&8&\\x_1&+5x_2&-x_3&=&4&\\-x_1&+11x_2&-4x_3&=&1&\end{array}\right

First, state the problem in matrix form, this means, extracting only the numbers, and putting them in a box.

\left[ \begin{array}{ccc|c} -3 & 1 & -2 & 8 \\\\ 1 & 5 & -3 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]

This is called an augmented matrix. The word “augmented” refers to the vertical line, which we draw to remind ourselves where the equals sign belong

Next, transform the augmented matrix to the reduced row echelon form with the help of Row Operations.

Row Operation 1: multiply the 1st row by -1/3

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 1 & 5 & -1 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]

Row Operation 2: add -1 times the 1st row to the 2nd row

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ -1 & 11 & -4 & 1 \end{array} \right]

Row Operation 3: add 1 times the 1st row to the 3rd row

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ 0 & \frac{32}{3} & - \frac{10}{3} & - \frac{5}{3} \end{array} \right]

Row Operation 4: multiply the 2nd row by 3/16

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]

Row Operation 5: add -32/3 times the 2nd row to the 3rd row

\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3}  & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]

Row Operation 6: multiply the 3rd row by -1/15

\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3}  & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & 1 \end{array} \right]

Row Operation 7: add -5/4 times the 3rd row to the 2nd row

\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3}  & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]

Row Operation 8: add 8/3 times the 3rd row to the 1st row

\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3}  & \frac{2}{3} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]

Row Operation 9: add 1/3 times the 2nd row to the 1st row

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]

The reduced row echelon form of the augmented matrix is

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]

which corresponds to the system

\left\begin{array}{ccccc}x_1&&+\frac{9}{16} x_3&=&0&\\&1x_2&-\frac{5}{16}x_3&=&0&\\&&0&=&1&\end{array}\right

Equation 3 cannot be solved, therefore, the system has no solution.

7 0
4 years ago
Given that f(x) = -x and g(x) = 2x - 1. Find the value of f(3) - 9(-4).
Ad libitum [116K]

Answer:

f(3)= -3

g(-4)=2*(-4 ) -1 =-9

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
HELPPP IM ALMOST DONE WITH THIS CLASS
iris [78.8K]
<h3>Answer: Choice D)  y = cos(2x)</h3>

Explanation:

Normally, cosine has a period of 2pi. This means that the curve repeats itself every 2pi units. However, this graph has a period of pi.

We can see this by noting that the distance from one peak such as x = 0 to another adjacent peak x = pi is exactly pi units across.

Since T = pi is the period, we then know that

B = (2pi)/T

B = (2pi)/(pi)

B = 2

Then recall that the general template is y = Acos(B(x-C))+D

In this case, A = 1, C = 0 and D = 0. So all of this leads to y = cos(2x)

5 0
3 years ago
Sam ran 3 1/4 miles on Tuesday and then ran 2 5/6 miles on Wednesday. How many total miles did Sam run?
Masteriza [31]

Sam run for total 6.08 miles.

Sam run 3 1/4 miles on Tuesday and then ran 2 5/6 miles on Wednesday.

<h3 /><h3>What is fraction?</h3>

Fraction is ratio of integers. example- 7/8.

Sam run 3\frac{1}{4} miles on Tuesday.
on 2\frac{5}{6} on Wednesday
Total run =Run on tuesday + Run on wednesday

Total run = (3\frac{1}{4} )+(2\frac{5}{6} )

= 13/4 + 17/6

Total run = 6.08

Thus Sam run for total 6.08 miles

Learn more about fraction here:
brainly.com/question/10354322

#SPJ1

5 0
3 years ago
ROUND 43.0810194943 TO THE NEAREST TEN-THOUSANDTH HELP!
REY [17]
43.0810
Since the hundred thousandths is a 1 which is less than a 5, you have to round down.
3 0
4 years ago
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