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blsea [12.9K]
3 years ago
10

Can you help me solve this problem & write my answer in simplest radical form.

Mathematics
1 answer:
olga55 [171]3 years ago
3 0

Answer:

18

Step-by-step explanation:

30-60-90 triangles have a set of x values that you can plug your numbers into.

we know that the side length opposite to 30 is x

the side length opposite to 60 is xsqrt3

the side length opposite to 90 is 2x

from this we can gather that y=2x where x=9 therefore y=18

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I need help finding the answers someone plz help
Harlamova29_29 [7]

Happy New Year from MrBillDoesMath!

Answer:

Proof by ASA congruence postulate. See below

Discussion:

  Fact 1 :     angle A = angle T             (given)

  Fact 2:    The angles on both sides of point X are equal as vertical angles

                 are equal.

From these facts it follows that angle M = angle H (as all plane triangles have 180 degrees). Also  AM = TH  (given) so

   In the left triangle                          In the right triangle

  (angle M, side AM, angle A)  =      (angle H, side TH, angle T)


Hence the triangles have two congruent angles, and congruent sides included between the angles, so they are congruent by  ASA.


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8 0
3 years ago
Please solve equations by completing square
BlackZzzverrR [31]
x^2-4x+1=0\\
x^2-4x+4-3=0\\
(x-2)^2=3\\
|x-2|=\sqrt3\\
x-2=\sqrt3 \vee x-2=-\sqrt3\\
x=2+\sqrt3 \vee x=2-\sqrt3

x^2+6x-9=0\\
x^2+6x+9-18=0\\
(x+3)^2=18\\
|x+3|=\sqrt{18}=3\sqrt2\\
x+3=3\sqrt2 \vee x+3=-3\sqrt2\\
x=-3+3\sqrt2 \vee x=-3-3\sqrt2\\

x^2+20x+3=0\\
x^2+20x+100-97=0\\
(x+10)^2=97\\
|x+10|=\sqrt{97}\\
x+10=\sqrt{97} \vee x+10=-\sqrt{97}\\
x=-10+\sqrt{97} \vee x=-10-\sqrt{97}

x^2-2x-5=0\\
x^2-2x+1-6=0\\
(x-1)^2=6\\
|x-1|=\sqrt6\\
x-1=\sqrt6 \vee x-1=-\sqrt6\\
x=1+\sqrt6 \vee x=1-\sqrt6
8 0
3 years ago
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