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4 years ago

Order the equations from least to greatest based on the number of solutions to each equation.

Mathematics

1 answer:

aleksandrvk [35]4 years ago

3 0

In order it goes like this

- 4^ x - 1 = 3^ (-x) - 2

- 3 x + 6 = 2^ x + 1

3^ x - 3 = 2 x - 2

so the question is the answer

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For this item, complete the choice matrix by clicking the appropriate answer in each row.

Vesna [10]

That person is really smart good job

8 0

3 years ago

Read 2 more answers

Two containers are mathematically similar. Their volumes are 54cm³ and 128cm³. The height of the smaller container is 4.5cm. Cal

Ivenika [448]

Answer:

the height of the larger container is 6 cm

Step-by-step explanation:

The computation of the height of the large container is shown below:

In the case when the factor of linear scale for two solid be k

So the factor of volume scale be k^3

Now the volume scale factor is

= 128 ÷ 54

= 64 ÷ 27

= (4 ÷ 3)^3

Now the height of the large container is

= 4.5 cm × 4 ÷ 3

= 6 cm

hence, the height of the larger container is 6 cm

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=3%20%5Ctimes%204x%20%5Ctimes%202y" id="TexFormula1" title="3 \times 4x \times 2y" alt="3 \time

Dahasolnce [82]

Answer:

24 x y

Step-by-step explanation:

Simplify the following:

3×4×2 x y

Hint: | Multiply 3 and 4 together.

3×4 = 12:

12×2 x y

Hint: | Multiply 12 and 2 together.

12×2 = 24:

Answer: 24 x y

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