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Yuri [45]
3 years ago
8

A circle is shown. Secants E G and D G intersect at point G outside of the circle. Secant E G intersects the circle at point F a

nd secant D G intersects the circle at point H. The length of E F is x, the length of F G is 6, the length of D H is x + 3, and the length of H G is 5. What is the length of line segment DG? 4 units 7 units 12 units 23 units
Mathematics
2 answers:
USPshnik [31]3 years ago
7 0

Answer:

Step-by-step explanation:

The formula we need here is

6(6+x)=5(5+x+3)

which simplifies to

6(6+x)=5(8+x)

which simplifies to

36 + 6x = 40 + 5x and

x = 4

So DG = 5 + 4 + 3 which is 12

pentagon [3]3 years ago
3 0

Answer:

The correct answer is C. 12

Step-by-step explanation:

I just took the test

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Answer:

yr point is A

Step-by-step explanation:

If a point is on the bisector of an Angle , then it is equidistant from two sides .

The Meaning of Bisector is to divide something or anything into two equal Halves.

The Meaning of Bisector of any Angle means to divide that Angle into two equal Parts.

Option A

equidistant from the two sides of the angle.

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3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
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Please help ASAP IM IN DESPREATE NEED OF HELP
Advocard [28]

Answer:

See the attached image for the graph of the first system.

Step-by-step explanation:

Here's how to graph the first system.

Start with the inequality -y \le -2x-3.  You can make this easier to work with by multiplying through by -1.  Remember to switch the inequality sign when multiplying by a <u>negative</u> number.  OK, you get the inequality

y \ge 2x+3.

The graph will be a half-plane -- all the points on one side of a line.  The line that is the boundary of the half-plane has an equation:  y=2x+3 -- just use an  =  sign instead of the inequality sign.

Graph the line.

The equation of the line is in slope-intercept form:  y = mx + b, so you can tell the y-intercept is 3 and the slope is 2 (think of it as a fraction 2/1).  Graph the line by going to the point (0, 3) -- the y-intercept -- then use the slope 2/1 interpreted as "rise over run" to go up 2 units and right 1 unit, arriving at the point (1, 5).  Draw the line through those points, (0, 3) and (1, 5).

Now you have to decide which side of the line the inequality y \ge 2x+3 is describing. To do this, pick a point which is not on the line, plug its coordinates into the inequality; if the result is true, shade the side of the line the point you picked is on (if false, shade the <u>other</u> side!)

An easy point to pick in this case is the origin, (0, 0).  Put zeros in for x and y in the inequality, and you'll get the statement 0 \ge 2(0)+3 \, \Rightarrow \, 0 \ge 3.  That's <u>false</u>, so shade the side of the line <u>not</u> containing the origin.  In the image below, the shading is in purple.

All right, now for the other inequality, x+2\le 0.  Subtract 2 from both sides and the inequality becomes x \le -2.  This, too, graphs as a half-plane whose boundary line has equation x=-2.  Graph the line.  A line with an equation that has  x  in it but not  y is a vertical line with all its x-coordinates equal to the number on the right side of the equation.  This line is vertical and goes through points such as (-2, 0).

Pick a point <u>not</u> on the line (the origin works again).  Put the coordinates into the inequality to get 0\le -2 which is <u>false</u>.  Shade the side of the vertical line which does <u>not</u> contain the origin.  In the image below, the shading is in black.

Finally,  YAY!  \o/ ,  the solutions to the system are all the points in the plane that got shaded twice.  In the image, they are the cross-hatched points above the purple line and to left of the black line.

Note: If you get a system with three inequalities, you'll be graphing three half-planes and looking for points that got shaded three times!

Note: One of your questions has the inequalities x \ge 0 and y \ge 0 in it.  These two inequalities say that the x and y coordinates are both positive or zero, confining your attention to Quadrant I in the upper-right part of a graph, above the x-axis <u>and</u> to the right of the y-axis.

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Sonja [21]
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Answer:

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Step-by-step explanation:

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