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Drupady [299]
3 years ago
8

What is the perimeter of a square if the length of one of the diagonals is 12 cm?

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
7 0

Answer:

Step-by-step explanation:

Use Pythagorean theorem to find the length of the square.

Length of the square = a

a² +a² = diagonal²

2a² = 12²

2a² = 144

a² = 144 ÷ 2

a² = 72

a =\sqrt{72}\\\\  = \sqrt{2*2*2*3*3}\\\\ = 2*3\sqrt{2}\\\\a = 6\sqrt{2}\\\\\\\\Perimeter= 4a\\\\ = 4*6\sqrt{2}\\\\= 24\sqrt{2} cm

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(1 point) Let y=ex/5. Find the differential dy when x=2 and dx=0.2 Find the differential dy when x=2 and dx=0.01
Deffense [45]

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Differential Calculus.

Since, we know this is a definite calculus,

thus we get here as, by using the Chain Rule.

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3 years ago
How do I differentiate this?​
disa [49]

Answer:

\frac{8}{\left(-x+2\right)^2} & x = 1, x = 3

Step-by-step explanation:

I'm sure you are familiar with the product rule,

If y = u*v => dy/dx = u * dv/dx + v * dy/dx <----- product rule

In this case:

u=3x+2,\:v=\left(2-x\right)^{-1},\\=>\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)

Now remember the sum rule:

\frac{d}{dx}\left(3x+2\right) = \frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(2\right),\\\frac{d}{dx}\left(3x\right) = 3,\\\frac{d}{dx}\left(2\right)  = 0\\\frac{d}{dx}\left(3x+2\right) = 3

For this second bit we apply the chain rule:

\frac{d}{dx}\left(\left(2-x\right)^{-1}\right) = -\frac{1}{\left(2-x\right)^2}\frac{d}{dx}\left(2-x\right),\\\frac{d}{dx}\left(2-x\right) = -1,\\\\=> -\frac{1}{\left(2-x\right)^2}\left(-1\right)\\=> \frac{1}{\left(2-x\right)^2}

If we substitute these values back into the expression...\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)

...we get the following:

3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)

The rest is just pure simplification:

3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)\\= \frac{3}{-x+2}+\frac{3x+2}{\left(-x+2\right)^2}\\= \frac{3\left(-x+2\right)}{\left(-x+2\right)^2}+\frac{3x+2}{\left(-x+2\right)^2}\\\\= \frac{3\left(-x+2\right)+3x+2}{\left(-x+2\right)^2}\\\\= \frac{8}{\left(-x+2\right)^2}

Now let's equate this to equal 8 for the second bit and solve for x:

\frac{8}{\left(-x+2\right)^2}=8,\\\frac{8}{\left(-x+2\right)^2}\left(-x+2\right)^2=8\left(-x+2\right)^2,\\8=8\left(-x+2\right)^2,\\\left(-x+2\right)^2=1,\\x = 1, x = 3

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The slope would be 2/3!
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Check the picture below.

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3 years ago
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Answer:

Hey! Ok so the answer is 6.

Step-by-step explanation:

If 1 pack has 8 bracelets in each and she has 48 in total then you would have to divide the total to the number of bracelets in each pack so therefore 48 divided by 8 which is 6

Hoped this helped! Have an amazing day! Im Eve btw. also consider marking this brainliest thank you so much if you do!✨

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