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3 years ago

Simplify (ignore my garbage camera quality ....)

Mathematics

2 answers:

AlexFokin [52]3 years ago

6 0

Answer:

The correct answer is option C

Step-by-step explanation:

Points to remember

Identities

(xᵃ)ᵇ = xᵃᵇ

(x¹/ᵃ)ᵇ = xᵃ/ᵇ

To find the correct answer

It is given that, (3¹/⁷)⁷

By using above identities we can write,

(3¹/⁷)⁷ = 3⁽¹/⁷ *⁷⁾

= 3¹

= 3

Therefore the simplified form of (3¹/⁷)⁷ = 3

The correct answer is option C

(3¹/⁷)⁷ = 3

monitta3 years ago

5 0

Answer:

C. 3

Step-by-step explanation:

Given expression is:

$({3^{\frac{1}{7}})^7$

We know that the rules of exponents are used to solve these kind of questions.

When there is exponent on exponent like in this question 1/7 has an exponent of 7 , the exponents are multiplied.

So,

$=3^{(7*\frac{1}{7} )}$

The 7's will be cancelled out and remaining power will be 1

$=3^1\\=3$

Hence, option C is correct ..

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There are 15 girls and 25 boys in a science club what percent of the members were girls

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15 / 40 = .375
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.375 = 37.5%

15/40 as a fraction ---> simplify ---> 3/8

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3 years ago

Which expression is not equivalent to the other expressions?

Ket [755]

That would be D

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3 years ago

Read 2 more answers

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

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