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melomori [17]
3 years ago
7

Please help me out on some questions...12th grade math..Pleasee help anyone​

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

If all the angles of the triangle must equal 180, then we have to subtract the 2 known values from 180.

180 - 68.5 - 49.5 = x

62 = x

So that means the missing side is equal to 62!

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

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ddd [48]
3996 + 7899 = 11895

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Flying against the wind, an airplane travels 4500km in 6 hours. Flying with the wind, the same plane travels 2910 kilometers in
katen-ka-za [31]

Let <em>a</em> denote the airplane's speed in still air, and <em>w</em> the windspeed.

When the plane flies against the wind, it can travel an average speed of

(4500 km) / (6 h) = 750 km/h

so that

<em>a</em> - <em>w</em> = 750 km/h

Flying with the wind, it moves at a speed of

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so that

<em>a</em> + <em>w</em> = 970 km/h

Add the two equations to eliminate <em>w</em> and solve for <em>a</em> :

(<em>a</em> - <em>w</em>) + (<em>a</em> + <em>w</em>) = 750 km/h + 970 km/h

2<em>a</em> = 1720 km/h

<em>a</em> = 860 km/h

Subtract them to eliminate <em>a</em> and solve for <em>w</em> :

(<em>a</em> - <em>w</em>) - (<em>a</em> + <em>w</em>) = 750 km/h - 970 km/h

-2<em>w</em> = -220 km/h

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7 0
3 years ago
Help me please LOL BAKANAJJAJAJANAJAHAHAHAHHAHA
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Prove that the diagonals of a parallelogram bisect each other​
Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof :\begin{array}{ |c| c |  c |  } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In  \: \triangle ^{s}  \:AOB \: and \: COD  } \\  \sf{(i)}&  \sf{ \angle \: OAB =  \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\  \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\  \sf{(iii)} &\sf{ \angle \: OBA=  \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\  \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}.          Proved ✔

♕ And we're done! Hurrayyy! ;)

# STUDY HARD! So, Tomorrow you can answer people like this , " Dude , I just bought this expensive mobile phone but it is not that expensive for me" [ - Unknown ] :P

☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

\underbrace{ \overbrace  {\mathfrak{Carry \: On \: Learning}}} ☂

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5 0
3 years ago
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