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Maslowich
3 years ago
8

Which equation shows a valid step in solving:

Mathematics
2 answers:
Volgvan3 years ago
8 0

Answer:

The equation (\sqrt[3]{2x-6})^3=(-\sqrt[3]{2x+6})^3 represents the valid step in solving the given expression \sqrt[3]{2x-6}+\sqrt[3]{2x+6}

Therefore the option (\sqrt[3]{2x-6})^3=(-\sqrt[3]{2x+6})^3

Step-by-step explanation:

Given expression is \sqrt[3]{2x-6}+\sqrt[3]{2x+6}

To solve the given expression as below :

\sqrt[3]{2x-6}+\sqrt[3]{2x+6}

\sqrt[3]{2x-6}=-\sqrt[3]{2x+6}

cubing on both sides we get

(\sqrt[3]{2x-6})^3=(-\sqrt[3]{2x+6})^3

The equation (\sqrt[3]{2x-6})^3=(-\sqrt[3]{2x+6})^3 represents the valid step in solving the given expression \sqrt[3]{2x-6}+\sqrt[3]{2x+6}

Therefore the option (\sqrt[3]{2x-6})^3=(-\sqrt[3]{2x+6})^3

olasank [31]3 years ago
8 0

Answer:

D

Step-by-step explanation:

e d g e n u i t y

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Anna007 [38]

1)

n         1          2         3           4         5         6

f(n)    1033    932     831      730     629     528

First term (a₁):  <u>1033 </u>          Common difference (d): <u>-101 </u>

Explicit rule:  a_{n} = 1134 - 101n    Recursive rule: a_{n} = a_{n-1} - 101

a_{n} = a_{1} + d(n - 1)

a_{n} = 1033 - 101(n - 1)

a_{n} = 1033 - 101n + 101

a_{n} = 1134 - 101n

***********************************************************************************

2)

n         1          2         3           4         5         6

f(n)   -39      -29       -19        -9          9        19

First term (a₁):  <u> -39  </u>          Common difference (d): <u> +10  </u>

Explicit rule:  a_{n} = -49 + 10n    Recursive rule: a_{n} = a_{n-1} + 10

a_{n} = a_{1} + d(n - 1)

a_{n} = -39 + 10(n - 1)

a_{n} = -39 + 10n - 10

a_{n} = -49 + 10n

***********************************************************************************

3)

n         1          2         3           4         5         6

f(n)   3.75      2.5     1.25        0      -1.25     -2.5

First term (a₁):  <u> 3.75  </u>          Common difference (d): <u> -1.25  </u>

Explicit rule:  a_{n} = 5 - 1.25n    Recursive rule: a_{n} = a_{n-1} - 1.25

a_{n} = a_{1} + d(n - 1)

a_{n} = 3.75 - 1.25(n - 1)

a_{n} = 3.75 - 1.25n + 1.25

a_{n} = 5 - 1.25n


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