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VMariaS [17]
3 years ago
5

I need to match each one to each other

Mathematics
1 answer:
kotegsom [21]3 years ago
4 0

Answer

speed : metres per second (a)

perimeter : meters (b)

volume : cubic centimetres (d)

area : square metres (c)

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Math help!!!!! thanks!!!!!!!
astraxan [27]
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We start with exponents.
4 x 4 is 16
12 divided by 3 is 4
4 times 5 is 20
20-16 is 4
4 is the answer.
5 0
3 years ago
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Please help me with this question
gladu [14]
The correct answer choice is (D) 1.10/1 Ib
Hope this helps!
5 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}}} ☂

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5 0
3 years ago
In most microcomputers the addresses of memory locations are specified in hexadecimal. These addresses are sequential numbers th
iren [92.7K]

Considering that the addresses of memory locations are specified in hexadecimal.

a) The number of memory locations in a memory address range ( 0000₁₆ to FFFF₁₆ )  = 65536 memory locations

b) The range of hex addresses in a microcomputer with 4096 memory locations is ;  4095

<u>applying the given data </u>:

a) first step : convert FFFF₁₆ to decimal           ( note F₁₆ = 15 decimal )

( F * 16^3 ) + ( F * 16^2 ) + ( F * 16^1 ) + ( F * 16^0 )

= ( 15 * 16^3 ) + ( 15 * 16^2 ) + ( 15 * 16^1 ) + ( 15 * 1 )

=  61440 + 3840 + 240 + 15 = 65535

∴ the memory locations from  0000₁₆ to FFFF₁₆ = from 0 to 65535 = 65536 locations

b) The range of hex addresses with a memory location of 4096

= 0000₁₆ to FFFF₁₆ =  0 to 4096

∴ the range = 4095

Hence we can conclude that the memory locations in ( a ) = 65536 while the range of hex addresses with a memory location of 4096 = 4095.

Learn more : brainly.com/question/18993173

6 0
3 years ago
1. Complete the inequality below to describe the
dezoksy [38]

Domain is the entire span left to right (on the x-axis) that the graph is on. Since the graph goes from x=-4 and ends at x=4, the domain would be from -4 to 4. The circle at -4 is open, so it does not include the point at -4, just everything leading up to it. So, the domain would be

- 4 < x \leqslant 4

The range is similar, it is the entire span that the graph goes up and down (on the y-axis). The graph starts at the bottom at y=-2, and ends at y=5. The bottom point (4,-2) is closed, so the graph includes that point, and the top point (-4,5) is open and doesn't include the point. Therefore, the range would be

- 2 \leqslant x < 5

5 0
3 years ago
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