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3 years ago

Consider the system of equations.

Mathematics

1 answer:

Ludmilka [50]3 years ago

4 0

then both equation r same after we re arrange it so no result

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Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

Andre45 [30]

Answer:

Option d) 5 to the power of negative 5 over 6 is correct.

$\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

ie, $5^\frac{\bf -5}{\bf 6}$

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

7 0

4 years ago

Does anyone know that 6 squared is I always forget

irakobra [83]

6 squared is 36. 6*6=36

7 0

4 years ago

Read 2 more answers

I need help on these questions

Kay [80]

1.) 10c - 12 = -2
-2C is the temperature in Tallahassee

2.) 48 kilometers divided the average of 3.2 Km
48÷3.2=15
15 is the value of D

3.) You seem to have 3 =D

4.) I'm pretty sure you just add $\frac{1}{4} + \frac{1}{3} = \frac{7}{12}$
7/12 is your answer

----------------------------------------------------------------------------------------------------------------
Thank You! Have a great day =D
and please correct me if wrong =D

7 0

3 years ago

What is the answer to this problem:

liraira [26]

Answer:

10.45

Step-by-step explanation: your welcome 4.75 X 2.2

6 0

3 years ago

Read 2 more answers

A couple is sitting inside a marriage hall, in a row that can accommodate a maximum of six couples. On the right of this couple

NeX [460]

There are 5 couples are sitting in the row

<h3>How many couples are sitting in the row?.</h3>

The given parameters are

Maximum = 6 couples

In the hall, we have the following arrangements

<u>First couple</u>

Right = 2 couples

Left = 1 couple

<u>Another couple</u>

Left = 3 couple

Right = 1 couples

The left of the other couple is the right of the first couple

So, we have the following arrangements:

<u>First couple (1)</u>

Right = 2 couples

Left = 1 couple

<u>Another couple</u>

Right = 1 couples

The total number of couples is

Total = 1 + 2 + 1 + 1

Evaluate the sum

Total = 5

Hence, there are 5 couples are sitting in the row