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dedylja [7]
3 years ago
14

The sum of two numbers is 6 The difference of the two numbers is 3 What are the two numbers.

Mathematics
2 answers:
riadik2000 [5.3K]3 years ago
4 0

Answer:

4.5 and 1.5

Step-by-step explanation:

Let the 2 numbers be x and y with x > y , then

x + y = 6 → (1)

x - y = 3 → (2)

Add the 2 equations term by term to eliminate y, that is

2x + 0 = 9

2x = 9 ( divide both sides by 2 )

x = 4.5

Substitute x = 4.5 into (1) for value of y

4.5 + y = 6 ( subtract 4.5 from both sides )

y = 1.5

The 2 numbers are 4.5 and 1.5

stiv31 [10]3 years ago
3 0

Answer:

4.5 and 1.5

Step-by-step explanation:

4.5+1.5=6

4.5-1.5=3(difference)

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zvonat [6]

Answer:

<h3>YOUR ANSWER IS IN THE ATTACHMENT!! </h3>

<h2>HOPE IT HELPS U!!!! </h2>

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Dy/dx = 2xy^2 and y(-1) = 2 find y(2)
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2887301

—————

Solve the initial value problem:

   dy
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\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\&#10;\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\&#10;\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\&#10;\mathsf{-\,\dfrac{1}{y}=x^2+C_1}

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Take the reciprocal of both sides, and then you have

\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}


In order to find the value of  C₁  , just plug in the equation above those known values for  x  and  y, then solve it for  C₁:

y = 2,  when  x = – 1. So,

\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\&#10;\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\&#10;\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\&#10;\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\&#10;\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}

\mathsf{C_1=-\,\dfrac{3}{2}}


Substitute that for  C₁  into (i), and you have

\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\&#10;\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\&#10;\mathsf{y=-\,\dfrac{2}{2x^2-3}}


So  y(– 2)  is

\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\&#10;\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\&#10;\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\&#10;\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}


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Tags:  <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0
3 years ago
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