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WINSTONCH [101]
3 years ago
14

To find the vertex, What must you do with the value you get from the formula −b/2a

Mathematics
1 answer:
PtichkaEL [24]3 years ago
8 0

Answer:

C.

Step-by-step explanation:

You would need to plug it into your example equation. This is because you would need the slope of that equation. Hope this helped you out!!!

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Which of the following fall(s) at a z-score of 0.0 in a normal distribution? Mean Median Mode Mean, media and mode
Artist 52 [7]

Answer: Median Mode Mean

Step-by-step explanation:

A normal distribution a symmetric distribution where most of the values lies around the central peak where Mean, median mode all lies together and the value of z=0 (as z-value given the distance of data values from mean with respect to the standard deviation).

i.e.  Z-score of 0 has 50% of the area to the left and 50% area to the right.

Hence, at a z-score of 0.0 in a normal distribution Mean, Median and Mode all fall together.

7 0
3 years ago
Annie and him are taking turns driving on their trip. On one day they drove a total of 420 miles. Annie drove 3 times farther th
JulijaS [17]
1,260 miles all you have to do is 420x3 because Annie drove 3 times faster.
3 0
3 years ago
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}}} ☂

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0
3 years ago
Right triangle FGH has midsegments of 15,36, and 39 centimeters given that area of a triangle is represented by the formula A=1/
Schach [20]

Answer:

Step-by-step explanation:

according to the midsegment theorem, a midsegment of a triangle is parallel to a side of a triangle and its length is 1/2 length of the side.

1/2FG= 15, SO FG=30,

1/2 GH=36 SO GH=72

AREA = 1/2(72*30)=1080 CM^2

3 0
3 years ago
Read 2 more answers
Can someone help me with this math question? The answer is suppose to be
Tanzania [10]

<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em>

3 0
3 years ago
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