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Komok [63]
3 years ago
8

X+7=2y+1 3x-y=7 solve for x and y

Mathematics
1 answer:
vovangra [49]3 years ago
6 0

{x+7=2y+1    ⇔  x-2y=1-7   ⇔ x-2y=-6

{3x-y=7     /*(-2)

{x-2y=-6

<u>{-6x+2y=-14      </u>(+)

<em>x-6x-2y+2y=-6-14</em>

<em>-5x=-20     </em>/:(-5)

x=4

x -2y=-6

4-2y=-6       /-4

-2y=-10     /:(-2)

y=5

<em>Solution: (4; 5)</em>



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4 0
3 years ago
30 POINTS!!! Kent won 25% of the games he played on Saturday. If he lost 12 games on Saturday, and no game ended in a tie, how m
Lana71 [14]

Answer:

He won 4 games

Step-by-step explanation:

Percentage of games he lost= 75%

Let the total number of games be x

75% of x = 12

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x= \frac{1200}{75}

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4 0
2 years ago
Find angle D if angle B = 50
Mariana [72]
<h3>Answer:  80 degrees</h3>

============================================================

Explanation:

I'm assuming that segments AD and CD are tangents to the circle.

We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.

By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.

----------------------------

Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.

Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.

We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.

---------------------------

Here's what we have so far for quadrilateral DAEC

  • angle A = 90
  • angle E = 100
  • angle C = 90
  • angle D = unknown

Now we'll use the idea that all four angles of any quadrilateral always add to 360 degrees

A+E+C+D = 360

90+100+90+D = 360

D+280 = 360

D = 360-280

D = 80

Or a shortcut you can take is to realize that angles E and D are supplementary

E+D = 180

100+D = 180

D = 180-100

D = 80

This only works if AD and CD are tangents.

Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.

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Olenka [21]
No because the stupidity of you can not calculate
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