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3 years ago

How would you read this function notation? f(x)=3x+1

Mathematics

2 answers:

mr Goodwill [35]3 years ago

8 0

You would read this as, "F of x equals 3x plus 1."

If this isn't the answer you are looking for, please let me know :>

Nonamiya [84]3 years ago

5 0

Answer (i think) f(x−1)−f(2)=3x−9

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Qhat is the ratio of 6to 5

asambeis [7]

6:5 is the ratio #answerwithquality #BAL

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3 years ago

A bottle holds 540 milliliter a of liquid. How much is this in ounces? Use the following conversion: 1punce is30 milliliter

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540mL/30mL=18 ounces

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4 years ago

In a square, doubling every side increases the area by 27 units. What is the length of side of the newly formed square? a) 3 b)

MatroZZZ [7]

<h3>Answer: B) 6</h3>

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Explanation:

x = original side length

2x = double the side length

The old area is x^2. The new square area is (2x)^2 = 4x^2

new area = (old area) + 27

4x^2 = x^2 + 27

4x^2-x^2 = 27

3x^2 = 27

x^2 = 27/3

x^2 = 9

x = sqrt(9)

x = 3

The old original square has a side length of 3 units.

The new larger square has a side length of 2x = 2*3 = 6 units which is the final answer (choice B)

old area = 3^2 = 9

new area = 6^2 = 36

The jump from 9 to 36 is +27 to help confirm the answer.

6 0

2 years ago

Read 2 more answers

Use a matrix to find the coordinates of the endpoints or vertices of the image of each figure under the given reflection.

hjlf

The vertices of the original quadrilateral can be written in matrix form using the vertex matrix. The vertex matrix is
$\begin{pmatrix}-5&-1&-3&-7\\ 4&-1&-6&-3\end{pmatrix}$

To find the coordinates of the endpoints or vertices of the image of the given coordinate points reflected about the y-axis, we just need to multiply the transformation matrix by the vertex matrix. The transformation matrix for this particular problem is
$\begin{pmatrix}-1&0\\ 0&1\end{pmatrix}$

Multiplying the two matrices, we have
$\begin{pmatrix}-1&0\\ 0&1\end{pmatrix}\begin{pmatrix}-5&-1&-3&-7\\ 4&-1&-6&-3\end{pmatrix}=\begin{pmatrix}5&1&3&7\\ 4&-1&-6&-3\end{pmatrix}$

Therefore, the coordinates of the endpoints or vertices of the image are (5,4), (1,-1), (3, -6) and (7, -3).

7 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

lana [24]

The series and the sigma notations are $\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$ , $\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$ , $\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$ and $\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$

<h3>How to match each series with the equivalent series written in sigma notation?</h3>

To do this, we simply expand each sigma notation.

So, we have:

$\sum\limits^4_0 3(5)^n$

Next, we set n = 0 to 4.

So, we have:

3(5)^0 = 3

3(5)^1 = 15

3(5)^2 = 75

3(5)^3 = 375

3(5)^4 = 1875

So, we have:

$\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$

$\sum\limits^4_0 4(8)^n$

Next, we set n = 0 to 4.

So, we have:

4(8)^0 = 4

4(8)^1 = 32

4(8)^2 = 256

4(8)^3 = 2048

4(8)^4 = 16384

So, we have:

$\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$

$\sum\limits^4_0 2(3)^n$

Next, we set n = 0 to 4.

So, we have:

2(3)^0 = 2

2(3)^1 = 6

2(3)^2 = 18

2(3)^3 = 54

2(3)^4 = 162

So, we have:

$\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$

$\sum\limits^4_0 5(3)^n$

Next, we set n = 0 to 4.

So, we have:

5(3)^0 = 5

5(3)^1 = 15

5(3)^2 = 45

5(3)^3 = 135

5(3)^4 = 405

$\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$

Hence, the series and the sigma notations are $\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$ , $\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$ , $\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$ and $\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$