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anyanavicka [17]
3 years ago
11

Find the point of intersection (if any) of the following pairs of lines: a) x – 2y +1=0 2x + 3y – 7 = 0 b) x-2y+11=0 -x+2y-13 =

0
Mathematics
1 answer:
pychu [463]3 years ago
5 0

Answer:

a) (11/7, 9/7)

b) There's no point of intersection

Step-by-step explanation:

a) x - 2y + 1 = 0

2x + 3y - 7 = 0

To find the point of intersection, we need to solve the system of equations  and the result will be the point of intersection (x,y)

x-2y+1=0\\x= 2y-1

Now we substitute x in the second equation:

2x+3y-7=0\\2(2y-1)+3y-7=0\\4y-2+3y-7=0\\7y-9=0\\y=9/7

Now we substitute y in our first equation.

x-2y+1=0\\x-2(9/7)+1=0\\x-18/7+1=0\\x=18/7-1\\x=11/7.

The point of intersection is (11/7, 9/7)

b) x -2y +11 =0

-x + 2y - 13 =0

We are going to follow the same procedure:

x-2y+11=0\\x=2y-11

-(2y-11)+2y-13=0\\-2y+11+2y-13=0\\0y=2\\0=2

Since this system of equations doesn't have a solution, the system has no point of intersection.

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The two curves y=x and y=x^{1/15} intersect at x=0 and x=1, with x^{1/15}\ge x over 0\le x\le1.

  • Washer method

\displaystyle\pi\int_0^1\left(\left(x^{1/15}\right)^2-x^2\right)\,\mathrm dx=\pi\int_0^1\left(x^{2/15}-x^2\right)\,\mathrm dx=\pi\left(\frac{15}{17}-\frac13\right)=\boxed{\frac{28\pi}{51}}

  • Shell method

We have y=x^{1/15}\implies x=y^{15}. The curves x=y and x=y^{15} intersect at y=0 and y=1, with y\ge y^{15} over 0\le y\le1.

\displaystyle2\pi\int_0^1y\left(y-y^{15}\right)\,\mathrm dy=2\pi\int_0^1\left(y^2-y^{16}\right)\,\mathrm dy=2\pi\left(\frac13-\frac1{17}\right)=\boxed{\frac{28\pi}{51}}

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3 years ago
In the triangle, find x, y, u, v ​
Gwar [14]

Answer:

  • x = 9
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  • v = 32

Step-by-step explanation:

Each of the triangles is similar, so we have ...

  3/5 = (3+x)/20

  x = 20(3/5) -3

  x = 9

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  4/5 = (4+y)/20

  y = 20(4/5) -4

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  3/5 = (3 + x + u)/60

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ch4aika [34]
Please refer to my image where it shows my work as I’m explaining.

Okay, for system 1:

1. I am using the elimination method to solve. So I check if all the terms are lined up and if any are the same. I found that 2X are common in both equations.

2. The goal is to “eliminate” the term hence the name. So I can choose to add or subtract. I chose subtraction because 2 - 2 equals 0 which is our goal. Solve for the rest of the terms. This will lead to getting y =4. Refer to image for the work.

3. Last step to to find the X value. We do this by picking any of the given equations,then substitute y with 4 and solve to eventually get x = 10. Refer to image for the work.

FOR SYSTEM 2:

1. Again, I am using the elimination method to solve. I noticed that NONE of the terms are in common so I will have to intervene. You can chose any term to create a match with but I chose Y since it was the one I could use the smallest number to multiply with. When multiplying, DONT just multiply Y, multiply ALL the terms in the equation or else everything will crash.

2. Now that I have terms in common I can choose to add or subtract. I chose subtraction because 2-2 equals zero which is what we want. Solve look at image for my process which lead to X = -8

3. Last step is to find the value of Y. Chose any of the given equations in system 2 then substitute x with -8. Refer to image to see process. It lead to y = 20


To check the validity of the answers, substitute the x and y values into both equations both side of the equal side should have the same number. Hope that helped!

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