After three cycles of successive approximation, without rounding the answers, the approximate x-value of the point of intersection is .

1a

The three lines are concurrent if they share a single point.

From the second equation,

x = 8y + 19

Substitute this value of x into the other equations.  This will allow you to solve for y.

2(8y + 19) + 3y = 0                              9(8y + 19) + 5y = 17

16y + 38 + 3y = 0                                 72y + 171 + 5y = 17

19y = -38                                                77y = -154

  y = -2                                                      y = -2

Notice that the two equations have the same y value as the solution.  Now we just evaluate the second equation to solve for x.  So we know that at some x value, the y-coordinate is -2.

x = 8(-2) + 19

x = -16 + 19

x = 3

The single point that these three equations share is (3, -2).

2a)

4x - 3y = 13                     eq1

-6x + 2y = -7                   eq2

Use the elimination method.  Multiply eq1 by 2 and multiply eq2 by 3.

8x - 6y = 26                eq1

-18 + 6y = -21              eq2

Add the equations to eliminate the y terms.

-10x = 5

   x = -1/2

Substitute this value of x into any of the equations to solve for y.

2b)

Substitute the first equation into the second equation.  This will get the second equation in terms of x.  Solve for x from that newly written equation.  Once you solved for x, substitute that value of x into the first equation to solve for y.  We did this method for the first question.

2c)

You have a vertical line that passes all points that have the x coordinate 7.

You have a horizontal line that passes all point that have the y coordinate -5.

If you were to graph these two lines, they will intersect at (7, -5).  Or you can just read the x and y values as (x, y).

3a)

Draw the following lines on a coordinate system:

i) A vertical line that passes through the points (3, 0).

ii) A horizontal line that passes through the point (0, 6).

iii) A line that passes through the points (0,0) and (1, -3).

Once you have drawn these lines throughout the coordinate system, look for 3 points of intersection.  Those points are the vertices of the triangle.  Be sure to use the line segments that make up that triangle.

3b)

By now, you should see that you get a right triangle.  Find the area of the two triangles using the formula

Area = (base × height) / 2

The length of the horizontal side is the base.  The length of the vertical side is the height.

4)

Lines that have the same slope never intersect.  Put both equations in  y=mx+b  form where the slope is the coefficient of x.

2x + 3y = 23  >  3y = -2x + 23   >   y = (-2 / 3)x + 23/3

7x + py = 8   >   py = -7x + 8     >   y = (-7 / p)x + 8/p

Set the slopes equal to each other.

-2 / 3 = -7 / p

Cross-multiply.

-2p = -21

Solve for p from this equation.   I leave this all to you.

Step-by-step explanation: maybe this will help you understand it

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

answer: it is d

step-by-step explanation: