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likoan [24]
3 years ago
5

2.25/295 = 4.85/x solve and show work

Mathematics
1 answer:
Marizza181 [45]3 years ago
7 0
2.25/295=4.85/x
multiply each side by 100
45/59=485/x
apply fraction cross multiply
45x=28615
divide 45 from each side
x=5723/9


I hope this helps!!!!!!!!
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Copy and complet the table compare the values of 2^x-2 with the values of 2^x-2
Vladimir79 [104]

#2^(x-2)

For 2

  • 2^(2-2)=2⁰=1

For 3

  • 2^(3-2)=2

For 4

  • 2^(4-2)=2²=4

For 5

  • 2^{5-2}=2³=8

For 6

  • 2^{6-2}=2⁴=16

#2^x-2

For 2

  • 2²-2=2

For 3

  • 2³-2=6

For 4

  • 2⁴-2=14

For 5

  • 2⁵-2=30

For 6

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3 0
2 years ago
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Solve st + 3t = 6 for s
vaieri [72.5K]

st + 3t = 6 for s

Subtract 3t to both sides

st + 3t  - 3t = 6 - 3t

Simplify

st = 6 - 3t

Divide both sides by t

st/t = (6-3t)/3

simplify

s = 6/3 - 3t/3

s = 2 - t


6 0
3 years ago
Help me preety please
maksim [4K]

Answer: $ 2.55

Step-by-step explanation:

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3 years ago
In a population of 10,000 geese the growth rate is 4% per year. What is the total increase in population over 4 years? Round you
omeli [17]

Answer:

Step-by-step explanation:

10000*4%= 400 Geese year1

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3 0
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Marcie bought a total of 20 used books and cds during a yard sale for a total of 54.50$. of books cost 1.50$ each and cds 5$ eac
melomori [17]
Let numbers of books be 'b' and numbers of CDs be 'c'

We can set up two equations:
Equation [1] ⇒ b+c=20
Equation [2] ⇒ 1.50b+5c=54.50

We are solving for the number of books and the number of CDs bought

When we have two equations in terms of two different variables; b and c, that we need to solve, then this becomes a simultaneous equation problem. 

First, rearrange Equation [1] to make either b or c the subject:
b+c=20
b=20-c

Then we substitute b=20-c into Equation [2]
1.50b+5c=54.50
1.50(20-c)+5c=54.50
30-1.50c+5c=54.50
5c-1.5c=54.50-30
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Now we know the value of c which is c=7, substitute this value into b=20-c we have b=20-7=13

Answer:
Numbers of books = 13
Numbers of CDs = 7
5 0
3 years ago
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