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Stels [109]
3 years ago
11

Julia examines the two triangles below and determines that .

Mathematics
2 answers:
baherus [9]3 years ago
8 0

IN Δ MLN:


∠M = 18.3 , ∠L = 98.6 AND ∠N = 180 - (∠M + ∠L) = 180 - (18.3 + 98.6 ) = 63.1



IN Δ FGH:


∠F = 98.6 , ∠G = 61.1 AND ∠H = 180 - (∠F + ∠G ) = 180 - (98.6 + 61.1 ) = 20.3



∴ ONLY ∠N = ∠F = 98.6


There is no other <span>congruent </span>angles


So, The correct statement is :

Inaccurate. The triangles are not similar because angle M is not congruent to angle H, and angle N is not congruent to angle G.



yulyashka [42]3 years ago
7 0

Answer:

It is "D" just took the quiz.

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<h2><em><u>You can solve this using the binomial probability formula.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2><em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
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\bold{\huge{\underline{ Solution }}}

<u>Before </u><u>answering </u><u>the </u><u>given </u><u>question</u><u>, </u><u>you </u><u>should </u><u>know </u><u>the </u><u>difference </u><u>between </u><u>polygon </u><u>and </u><u>regular </u><u>polygon</u><u>. </u>

  • Polygon :- It is closed figure constitute of straight line having different measurements and also having varies angles.
  • Regular polygon :- It is also a closed figure constitute of straight lines which having equal measuresment and all the angles of the given polygon are equal .

<u>There </u><u>are </u><u>different </u><u>types </u><u>of </u><u>polygon </u><u>based </u><u>on </u><u>the </u><u>sides </u><u>:</u><u>-</u>

  • A polygon having 3 sided called triangle ,having 4 sides called quadrilateral ,having 5 sides pentagon and so on.

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

I have thought to make a 4 side polygon that is quadrilateral.

<u>Steps </u><u>of </u><u>construction </u><u>:</u><u>-</u><u> </u>

  • Step - 1 :- Draw a 5 cm line AB horizontally
  • Step - 2 :- Then , From a point B, Draw a Vertical line of 5 cm and named it BC
  • Step - 3 :- Then, From a point C, Draw a horizontal line of 5 cm and named it CD
  • Step - 4 :- Now, join the line AD
  • Step - 5 :- Measure all the angles form by the 4 sides of the quadrilateral.

<u>Result </u><u>:</u><u>-</u><u> </u> All the angles are equal that is 90° each.

<u>Conclusion </u><u>:</u><u>-</u><u> </u> The above figure is square as it is having equal sides and angles.

<h3><u>Now</u><u>, </u><u>we </u><u>have </u><u>to </u><u>find </u><u>the </u><u>sum </u><u>of </u><u>the</u><u> </u><u>angles</u><u> </u><u>of </u><u>the </u><u>polygon </u><u>:</u><u>-</u></h3>

<u>Here</u><u>, </u><u> </u>

  • All angles are equal in measure 90°

<u>Therefore</u><u>, </u>

The sum of the angles of the given polygon

\sf{ = 90{\degree} + 90{\degree} + 90{\degree} + 90{\degree} }

\sf{ = 180{\degree} + 90{\degree} + 90{\degree} }

\sf{ = 270{\degree} + 90{\degree} }

\sf{ = 360{\degree} }

Hence, The sum of the angles of the above polygon is 360° .

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