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-BARSIC- [3]
1 year ago
6

What is the area of the figure below ​

Mathematics
1 answer:
aliya0001 [1]1 year ago
8 0

Answer:

Total Area = 68

Step-by-step explanation:

Comment

There is 1 square at each end.

There are 2 congruent trapezoids in the middle

Draw a 7 inch line across the widest part of the middle

Area of the squares

Area = s^2

s = 3

Area = 3^2

Area = 9

But there are 2 of them                                        18

Area of the 2 trapezoids

Area = (b1 + b2)*h / 2

b1 = 3

b2 = 7    Notice that 7 was the line you drew

h = 5

Area = (3 + 7)*5/2

Area = 50 / 2

Area = 25

But there are 2 of them                  <u>                      50</u>

Total Area                                                              68

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Can someone pls help w finding the x and y intercepts of this? <br><br> y=x^2-2x
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Answer:

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Step-by-step explanation:

For y-intercept, x=0 :

{ \tt{y =  {(0)}^{2} - 2(0) }} \\ { \tt{y = 0}}

For x-intercept, y=0 :

{ \tt{0 =  {2x}^{2} - 2x }} \\ { \tt{2x(x - 1) = 0}} \\ { \tt{x = 0 \:  \: and \:  \: 1}}

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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}\\\\&#10;\mathsf{y^{-2}\,dy=2x\,dx}


Integrate both sides:

\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}

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


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\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

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I hope this helps. =)


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

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