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Serggg [28]
3 years ago
9

Find The Quotient. 7/9 Divided By 2/3 Please Help

Mathematics
2 answers:
Flura [38]3 years ago
5 0
7/9 * 3/2 = 21/18 = 7/6 = 1 1/6
liq [111]3 years ago
5 0
7/9 ÷ 2/3
= 7/9 (3/2)
= 21/18

21 = 18q + r
clearly q=1 and r=3
therefore quotient is 1.
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7 0
3 years ago
Read 2 more answers
What is the surface area of the cylinder with height 7 cm and radius 5 cm? Round
nalin [4]

Answer:

376.991 cm²

Step-by-step explanation:

SA_{cylinder} = 2\pi r^{2}  + 2\pi rh

Surface Area = 2π(5²) + 2π(5)(7)

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                      = 120π

                      = 376.9911184... ≈ 376.991

<h2>Derive Surface Area of A Cylinder</h2>

If you take a look at a net of a cylinder, you can see it is composed of two circles(bases) and a rectangular strip with the length of the diameter of the circle

The formula for the area of a circle = πr², and since there are 2, it is 2πr²

The formula for circumference of a circle = 2πr, and since we are multiplying that by the height by the height of the cylinder, it is 2πrh

∴ 2πr² + 2πrh

5 0
2 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
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