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tia_tia [17]
3 years ago
13

Find three consecutive even integers such that the sum of the least integer and the middle integer is 3030 more than the greates

t integer.
Mathematics
1 answer:
Genrish500 [490]3 years ago
3 0
X = <span> least integer
x + 2 = </span><span>middle integer
x + 4 = </span><span>greatest integer

x + x + 2 - (x + 4) = 3030
2x + 2 - x - 4 = 3030
x - 2 = 3030
x = 3030 + 2  
x = 3032  </span>← least integer

middle integer = x + 2 = 3034

greatest integer = x + 4 = 3032 + 4 = 3036

Check:
3032 + 3034 = 6066
3036 + 3030 = 6066
6066 = 6066

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72x + 12y = 60 Which equation represents the same function? ​
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If three tangents to a circle form an equilateral triangle, prove that the tangent points form an equilateral triangle inscribed
weqwewe [10]
I added a figure so you can guide yourself throughout the proof I'm about to write, so I recommend that you download the picture beforehand and have this window and the picture's window open. Alright, let's get started!
Assuming that FED is an equilateral triangle according to the wording of the problem, we have that the angles \widehat{AFB}=\widehat{BEC}=\widehat{CDA}=60.
We also know that the circle in green is Inscribed in FED.

The following applies to every inscribed circle inside a triangle:
The center of the inscribed circle of a triangle is the intercept of all three angle bisectors of the triangle.

The above theorem implies that the line (FO) is an angle bisector because it goes through the vertex F and the center of the inscribed circle.

The previous statement implies that the angle \widehat{OFB}=30.

Now let's work on the OFB triangle.
Knowing that \widehat{OBF}=90 (because (EF) is tangent to the circle at B and OB is a radius of the circle. If you're lost here, remember that a tangent to a circle is always perpendicular to the radius of the circle.) we can then derive that \widehat{FOB}=180-90-30=60 (because the sum of the measures of all angles in a triangle is always equal to 180 degrees).

In the same way, we can prove also that:
\widehat{BOE}=\widehat{EOC}=\widehat{COD}=\widehat{DOA}=\widehat{AOF}=60 degrees.
Knowing the above we notice that \widehat{BOC}=\widehat{COA}=\widehat{AOB}=120degrees.

We're at the last part of our proof here:
Now notice that \widehat{BOA} subtends the same arc on the circle  that \widehat{BCA}.
According to the inscribed angle theorem, an angle \theta inscribed in a circle is half of the central angle 2\theta that subtends the same arc on the circle. 
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We can prove in a similar fashion that:  \widehat{CAB}=\widehat{ABC}=60degrees 
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5 0
2 years ago
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Gnesinka [82]

Answer:

to go from dilation S to dilation T

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ANSWER:

A.

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