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OverLord2011 [107]
3 years ago
13

Question 1

Mathematics
1 answer:
allochka39001 [22]3 years ago
5 0

Answer:

Step-by-step explanation:

Perimeter of quarter circle = r + r + \frac{1}{4}*2*\pi *r

            = 10 + 10 + \frac{1}{4}*2*3.14*10\\\\= 10 + 10 +  3.14*5\\\\\\= 10 + 10 + 15.70\\\\= 35.70 cm

2) d = 2*10 = 20 cm

Perimeter of semicircle = diameter + (1/2) * circumference of circle

                         = 20 +  \frac{1}{2}* 2*\pi *r\\\\= 20 + \frac{1}{2}*2*3.14*10\\\\= 20 + 3.14*10\\\\= 20 + 31.4\\\\= 51.40 cm

3) Perimeter of three-quarter =  r + r + (3/4)* circumference of circle

          = 10 + 10 + \frac{3}{4}*2*\pi *r\\\\= 10 + 10 + \frac{3}{4} * 2 * 3.14 * 10\\\\= 10 + 10 + 3 * 3.14 * 5\\\\= 10 + 10 + 47.1\\\\= 67.1 cm

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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
———  =  2xy²,      y = 2,  when x = – 1.
   dx


Separate the variables in the equation above:

\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}


Integrate both sides:

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

\mathsf{\dfrac{1}{y}=-(x^2+C_1)}


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}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\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}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}


So  y(– 2)  is

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


I hope this helps. =)


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

7 0
3 years ago
Find the solution for 2-x/3=5
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Answer:

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Step-by-step explanation:

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3 years ago
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Which multiplication expression is eqeul to 3/5 ÷ 1/2
eduard

Answer:

Lets solve = 3/5 ÷ 1/2

3/5 ÷ 1/2

= 3/5 × 2/1

= 3 × 2/5 × 1

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In mixed fraction its

1 \frac{1}{5}

____________________

Now lets solve "a" option

3/5 × 2

= 3 × 2/5 × 1

= 6/5

In mixed fraction

1 \frac{1}{5}

So option a = 3/5 x 2/1 is ur answer

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