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7 months ago

I need help what is 2,830 ÷ 28 =? Step-by-step pls

Mathematics

2 answers:

goldfiish [28.3K]7 months ago

7 0

Answer: 101 with a Remainder of 2

Step-by-step explanation:

vaieri [72.5K]7 months ago

4 0

2830 / 28 = 101
28
03
30
28
—-
02

Answer: 2830 / 28 = 101 with remainder of 2

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3 years ago

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Could somebody solve this for me step-by-step? Please!

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Answer: the answer to (g+h)x4 is

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Step-by-step explanation:

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2 years ago

For the differential equations dydx=sqrt(y^2−36) does the existence/uniqueness theorem guarantee that there is a solution to thi

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Answer:

1. (-4,6) there is no a solution to the equation through this point

2. (2,−6) there is no a solution to the equation through this point

3. (−5,39) there is a solution to the equation through this point

4. (−1,45) there is a solution to the equation through this point

Step-by-step explanation:

Using the existence and uniqueness theorem:

Let:

$F(x,y)=\sqrt{y^2-36} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{y}{\sqrt{y^2-36} }$

Now, let's find the domain of $F(x,y)$ , due to the square root:

$y^2-36 \geq 0\\\\y^2\geq 36\\\\y \geq 6\hspace{12}or\hspace{12}y\leq-6$

So the domain of the function is:

$y \in R\hspace{12}y\geq6\hspace{12}or\hspace{12}y\leq-6$

Now, due to the fraction $\frac{\partial F}{\partial y}$ the denominator must be also different from 0, so:

$y^2-36\neq0\\\\y \neq \pm6$

So, the theorem tells us that for each $y_0\in R:\hspace{12}y_0>6\hspace{12}or\hspace{12}y_0$ there exists a unique solution defined in an open interval around $x_0$ .

1. (-4,6) there is no a solution to the equation through this point because $y_0=6$

2. (2,−6) there is no a solution to the equation through this point because

$y_0=-6$

3. (−5,39) there is a solution to the equation through this point because

$y_0>6$

4. (−1,45) there is a solution to the equation through this point because

$y_0>6$

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