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4 years ago

What is −1/3−1 2/3=?

Mathematics

1 answer:

ra1l [238]4 years ago

7 0

The answer to <span>−1/3−1 2/3= is -2
ANSWER: </span><span>−1/3−1 2/3= -2

I hope I helped.
Have a good day.</span>

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A label wraps around soup can with a radius of 4.3 cm and a height of 6 cm. What is the lateral surface area of the label of the

Soloha48 [4]

Answer:

$52\pi \text{ cm}^2\approx 162.11\text{ cm}^2$

Step-by-step explanation:

We have been given that a label wraps around soup can with a radius of 4.3 cm and a height of 6 cm. We are asked to find the lateral surface area of the label of the soup can in square centimeters.

We will use lateral surface area of cylinder to solve our given problem as:

$LA=2\pi rh$ , where,

r = Radius of cylinder,

h = height of cylinder.

Upon substituting our given values in above formula, we will get:

$LA=2\pi \times\text{4.3 cm}\times \text{6 cm}$

$LA=51.6\pi \text{ cm}^2$

$LA=162.10618\text{ cm}^2$

Therefore, the lateral surface area of the label of the soup is approximately 162.11 square cm.

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4 years ago

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8_murik_8 [283]

I believe this is how you solve it

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3 years ago

Complete the following equation using <,>, or = 7___24/4

olga2289 [7]

Answer:

Step-by-step explanation:

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3 years ago

Can you help me with this question?-8(-4x-1)-9x

Contact [7]

First you have to multiply -8 to (-4x-1)
which is, -12x+8
then there is -9x you have to add like terms
-12x-9x+8
-21x+8 is your final answer

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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]

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

$
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3 years ago