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Tanya [424]
3 years ago
11

Sarah and Bryan went shopping and spend a total of $47.50. Bryan spends $15.50 less than was Sarah spent. How much did Bryan spe

nd?
Please help
Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
4 0
Hey there,
Sarah= $x
Bryan = $x - $15.50
$x + $x - $ 15.50 = $47.50
$x + $x = $47.50 + $15.50
$2x = $63
$x = $63 / 2 
     = $31.50
Bryan = $31.50 - $15.50
           = $16

Hope this helps :))

~Top
myrzilka [38]3 years ago
3 0
X+(x+15.50)=47.50
2x=32
x=16
Bryan spends $16
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In 2010, the population of a town was 8500. The population decreased by 4.5% each year.
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This is a problem in exponential decay.

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To solve for (number of years since 2010) we take logs of both sides

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-0.0843208857  / -0.0199966284  = number of years since 2010

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10) Show that in a group of 10 people (where any two people are either friends or enemies), there are either three mutual friend
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Answer with explanation:

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The combination between two people is that, they can be either friends or enemies.

Total number of Possible Relation

                            =18 +16+14+12+10+8+6+4+2

                            = 90 Relations in all.

Out of 90 relations , 45 will be friends and 45 will be enemies.

⇒Now, we have to prove that, between 10 people, there are either three mutual friends or four mutual enemies, and there are either three mutual enemies or four mutual friends.

→→If there are three mutual friends, total number of people in the group =3 ×2=6 people in the group

And, four mutual enemies in a group means there are 2 people in each group.

So, total number of people if we combine the two groups in which there are either three mutual friends or four mutual enemies

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Hence proved.

→→→→Second part is ,in this group there can be either three mutual enemies or four mutual friends.

⇒If there are three mutual enemies, total number of people in the group =3 ×2=6 people in the group

And, four mutual friends in a group means there are 2 people in each group.

So, total number of people if we combine the two groups in which there are either three mutual enemies or four mutual enemies

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Hence proved.

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3 years ago
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