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3 years ago

Two trains are traveling at constant speeds on different tracks.

Mathematics

1 answer:

Natasha2012 [34]3 years ago

3 0

Answer:

Train B is traveling faster

Step-by-step explanation:

From the picture, we want to know which train is traveling faster

We can know this by knowing the speed of the trains

Mathematically;

speed = distance/time

Train A

12.5 m in 1 second

speed = 12.5/1 = 12.5 m/s

Train B

100 m in 4 seconds

speed = 100/4 = 25 m/s

So train B is traveling faster

This is because it covered a distance of 100 m in 4 seconds as opposed to train A which covered a distance of 12.5 m

in 1 second

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Find the value of 100-98+96-94+92-90+...+8-6+4-2

Pachacha [2.7K]

Answer:

solution

100-98 =2

96-94 = 2

92-90 = 2

8-6 = 2

4-2 =2

2+2+2+2+2=10✓

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3 years ago

Find the length of segment AC if AB is 7, BC is 12 and B is between a A and C

Alex17521 [72]

Answer:

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3 years ago

MATH ANYONE PLEASE HELP

BartSMP [9]

3/8 is the answer to this question

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3 years ago

Read 2 more answers

Part 1: Create a scenario for an arithmetic sequence. For example, Jasmine practices the piano for ______ minutes on Monday. Eve

mr Goodwill [35]

Let $a_n$ be the n'th term of a sequence,

for example $a_1$ is the first term, $a_2$ is the second term and so on.
----------------------------------------------------------------------------------------------------
a sequence is arithmetic if the difference between any 2 terms is equal:

so an arithmetic sequence, for the first term = t, and the common difference = d, has the following form:

$a_1=t$

$a_2=t+d$

$a_3=t+d+d$

$a_4=t+d+d+d$

so clearly $a_n=t+d*(n-1)$
---------------------------------------------------------------------------------------------------
A sequence is geometric, if the ratio between any 2 consecutive terms is the same, and it is called the common ratio.

a geometric ratio with first term = t, and common ratio = r is:

$a_1=t$

$a_2=t*r$

$a_3=t*r*r$

$a_4=t*r*r*r$

thus clearly $a_n=t* r^{n-1}$
--------------------------------------------------------------------------------------------------

Part 1:

"<span>Jasmine practices the piano for __30___ minutes on Monday. Every day she ____increases_______ her practice time by ____5 minutes_____. </span>"

let $a_n$ be the number of hours Jasmine practices the n'th day,

for n=7, t=30, d=5 we have

$a_7=30+5*(7-1)=30+5*6=30+30=60$ ,

minutes is the time Jasmine practices on the 7the day.

Part 2

"<span>Anthony goes to the gym for ___60___ minutes on Monday. Every day he ___increases______his gym time by ____1/10_of the previous day_____. </span>"

if $S_n$ represents the minutes that Anthony goes to gym on the n'th day,

then

$S_1=60$

$S_2=60+1/10*60=10/10*60+1/10*60=11/10*60=1.1*60$

$S_3=1.1*1.1*60$

$S_4=1.1*1.1*1.1*60$

$S_5=1.1*1.1*1.1*1.1*60= (1.1)^{4}*60= 1.4641*60=87.85$
(minutes)

Part 3: it is clear that the formula that can be derived in Part 2 is:

$S_n=60*(1.1)^{n-1}$

so assume we want to find how many hours does Anthony spend in the gym on the 10th day,

then we calculate $S_10=60*(1.1)^{9}=60*2.358=141.5$ (minutes)

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3 years ago

Write 5000 = 12 as an order pair

mina [271]

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3 years ago