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3 years ago

Simplified form please help thank you

Mathematics

1 answer:

Agata [3.3K]3 years ago

7 0

$\dfrac{5^2}{2^{-2}}=5^2\cdot2^2=(5\cdot2)^2=10^2=100$

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What is the smallest subset that 46 belongs to

xeze [42]

Answer:

Natural/Counting Numbers

Step-by-step explanation:

It is the smallest subset a number can classify into. It is all of the numbers from 1 to positive infinity.

7 0

3 years ago

5. In this diagram, lines AC and DE are parallel,

zhannawk [14.2K]

The length of side DE is 3cm

<h2 /><h3>Parallel line</h3>

Since the line AC is equivalent to DE, hence AC = DE

To get the length of DE, we will need to get the length oF AC using the Pythagoras theorem:

Given the following parameters

Hypotenuse = 5

Opposite = 4

Required

Adjacent side

Substitute the given parameters into the formula of Pythagoras theorem.

AC² = 5² - 4²

AC² = 25 - 16

AC² = 9

AC = 3

Hence the length of side DE is 3cm

Learn more on Pythagoras theorem here: brainly.com/question/12306722

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3 years ago

The director of a museum recorded the number of students and adult chaperones in several school groups visiting the museum. She

sasho [114]

Answer:

b. 119

Step-by-step explanation:

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3 years ago

Read 2 more answers

What is x to the power of 3 divided by x to the power of -5

Veseljchak [2.6K]

X to the power of 8
When dividing powers, you subtract the "small numbers". When multiplying, you add them. 3-(-5) = 8

Hope it helps!

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3 years ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

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3 years ago