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4 years ago

Simplify (0.4) to the third power

Mathematics

2 answers:

svetoff [14.1K]4 years ago

7 0

0.064 is 0.4 to the third power

faust18 [17]4 years ago

5 0

Answer:

The simplified form of the provided expression is 0.064

Step-by-step explanation:

Consider the provided expression.

$(0.4)^3$

Now, use the property of exponent: $a^n=a\times a \times a.....n\ times$

Use the above property to solve the provided expression.

$0.4 \times 0.4 \times 0.4$

$0.064$

Thus, the simplified form of the provided expression is 0.064

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What is the Quotient of 1 and 2/3 divided by 2 and 3/5

Nostrana [21]

Answer:

25/39

Step-by-step explanation:

For division of mixed numbers, it often works well to first convert them to improper fractions. Then the division proceeds in the usual way: "invert and multiply", or "dot, swap."

$1\dfrac{2}{3}\div2\dfrac{3}{5}=\dfrac{5}{3}\div\dfrac{13}{5}=\dfrac{5}{3}\cdot\dfrac{5}{13}\\\\=\dfrac{5\cdot5}{3\cdot13}=\boxed{\dfrac{25}{39}}$

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<em>Additional comment</em>

Your graphing calculator can do this, too.

7 0

2 years ago

WILL GIVE BRAINLIEST! BUT YOU MUST EXPLAIN<br><br> explain clearly ...

aleksandr82 [10.1K]

Answer:

Step-by-step explanation:

Add all numbers then divide by 6 (Add all numbers including 94 then divide by six to get your answer)

5 0

3 years ago

Read 2 more answers

Mr. Gubareff gave his class 15 minutes to read. Sabira read 6 ¼ pages in that time. At what rate, in pages per hour, did Sabira

creativ13 [48]

Answer:

I think its 25 (proceed with caution)

Step-by-step explanation:

15×4=60 so basically 6.25×4=25

6.25 = 6 1/4

4 0

3 years ago

Read 2 more answers

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities