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Harlamova29_29 [7]

3 years ago

10

Write 2³ in simplified form

2 answers:

jenyasd209 [6]3 years ago

8 0

The answer is: 8
The 3 will represent how many times you need to multiple 2.
Example: 2x2x2=
First multiple 2x2=4
Then multiple 4x2= 8

DIA [1.3K]3 years ago

6 0

Answer:

8

Step-by-step explanation:

2^1 = 2×1 = 2

2^2= 2×2= 4

2^3= 2×2×2= 8

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Which of the following is NOT an EXPRESSION?<br> 4x²<br> X+4=8<br> 6010-4)<br> 14+ (-8x)

denis-greek [22]

Answer:

X+4=8

Step-by-step explanation:

X+4=8 has an equal sign. Equations have equal signs, not expressions.

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Anyone know this? Any help would be appreciated

zysi [14]

The correct answer is B.

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What is the mathematical induction of <br> 1+3+5+...+(2n-1)=n^2

Lena [83]

Let $S(n)$ be the statement that

$1+3+5+\cdots+(2n-1)=\displaystyle\sum_{k=1}^n(2k-1)=n^2$

Check to see that $S(n)$ holds for $n=1$ :

$\displaystyle\sum_{k=1}^1(2k-1)=1=1^2\implies S(1)\text{ is true}$

Now suppose that $S(N)$ is true, and use that assumption to show that $S(N+1)$ must also be true. When $n=N+1$ , you have

$\displaystyle\sum_{k=1}^{N+1}(2k-1)=\sum_{k=1}^N(2k-1)+2(N+1)-1$
$=N^2+2(N+1)-1$
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and so $S(N+1)$ is also true, thus proving the statement is true for all $n$ .

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4 years ago

First derivative of <br>√{cosec2x).show with full step.

Mice21 [21]

Answer:

$- \sf \displaystyle \: \frac{ \cos(2x) }{ \sin ^{2} (2x)\sqrt{ \csc(2x) } }$

Step-by-step explanation:

we are given a derivative

$\displaystyle \: \frac{d}{dx} ( \sqrt{ \csc(2x) } )$

and said to figure out the first derivative

to do so

recall chain rule:

$\sf\displaystyle \: \frac{d}{dx} (f(g(x)) = \frac{d}{dg} (f(g(x)) \times \frac{d}{dx} (g)$

so we get

$\displaystyle \: g(x) = \csc(2x)$

rewrite the derivative using the chain rule:

$\displaystyle \: \frac{d}{dg} ( \sqrt{ g } ) \times \frac{d}{dx} ( \csc(2x) )$

use square root derivative rule to simplify:

$\displaystyle \: \frac{1}{ 2\sqrt{g} } \times \frac{d}{dx} ( \csc(2x) )$

now we need to again use chain rule composite function derivative to simplify

where we'll take a new function n so we won't mess up two g's and we'll take 2x as n

use composite function derivative to simplify:

$\sf \displaystyle \: \frac{1}{ 2\sqrt{g} } \times \frac{d}{dn}( \csc(n) ) \times \frac{d}{dx} (2x)$

use derivative formula to simplify derivatives:

$\sf \displaystyle \: \frac{1}{ 2\sqrt{g} } \times - \cot(n) \csc(n) \times 2$

substitute the value of n:

$\sf \displaystyle \: \frac{1}{ 2\sqrt{g} } \times - 2\cot(2x) \csc(2x)$

substitute the value of g:

$\sf \displaystyle \: \frac{1}{ 2\sqrt{ \csc(2x) } } \times - 2\cot(2x) \csc(2x)$

now we need our trigonometric skills to simplify

rewrite cot and csc:

$\sf \displaystyle \: \frac{1}{ 2\sqrt{ \csc(2x) } } \times - 2 \dfrac{ \cos(2x) }{ \sin(2x) } \dfrac{1}{ \sin(2x) }$

simplify multiplication:

$\sf \displaystyle \: \frac{1}{ \cancel{ \: 2}\sqrt{ \csc(2x) } } \times \cancel{- 2} \dfrac{ \cos(2x) }{ \sin ^{2} (2x) }$

simplify multiplication:

$- \sf \displaystyle \: \frac{ \cos(2x) }{ \sin ^{2} (2x)\sqrt{ \csc(2x) } }$

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4 years ago

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