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zysi [14]
3 years ago
11

3.9 practice complete your assignment english 111

Mathematics
1 answer:
Citrus2011 [14]3 years ago
3 0
The answer is 15.9 because 3.9+111=15.9
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Write 3a^2 b^2 c^5 / 8x^4 y^3 z using no denominator
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If your starting salary is $40,000 and you receive a 3% increase at the end of every year, what is the total amount, in dollars,
sasho [114]

Answer:

SEE BELOW      PLEASE GIVE BRAINLIEST

Step-by-step explanation:

YEAR 1:  40000 X .03 = 1200       40,000 + 1200 = 41200

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YEAR 3: 42436 X .03 = 1273.08    42436 + 1273.08 = 43709.08

YEAR 4: 43709.08 x .03 = 1311.27    43709.08 + 1311.27 = 45020.35

YEAR 5: 45020.35 X .03 = 1350.61    45020.35 + 1350.61 = 46370.96

YEAR 6: 46370.96 X .03 = 1391.13         46370.96 + 1391.13 = 47762.09

YEAR 7: 47762.09 X .03 = 1432.86       47762.09 + 1432.86 = 49194.95

YEAR 8: 49194.95 X .03 = 1475.85        49194.95 + 1475.85 = 50670.80

YEAR 9: 50670.80 X .03 =  1520.12         50670.80 + 1520.12 = 52190.92

YEAR 10: 52190.92 X .03 = 1565.73         52190.92 + 1565.73 = 53756.65

YEAR 11: 53756.65 X .03 = 1612.70             53756.65 + 1612.70 = 55369.35

YEAR 12: 55369.35 X .03 = 1661.08           55369.35 + 1661.08 = 57030.43

YEAR 13: 57030.43 X .03 = 1710.91            57030.43 + 1710.91 = 58741.34

YEAR 14: 58741.34 X .03 = 1762.24            58741.34 + 1762.24 = 60503.58

YEAR 15: 60503.58 X .03 = 1815.11           60503.58 + 1815.11 = 62318.69

YEAR 16: 62318.69 X .03 = 1869.56         62318.69 + 1869.56 = $64,188.25 = $64188

3 0
3 years ago
Read 2 more answers
Whats 7 over 4 =35 over m in solving proportions NO LINKS PLZ
tatiyna

Hey there!

7/4 = 35/m

CROSS MULTIPLY BOTH of your NUMBERS

7(m) = 4(35)

7(m) = 7m

4(35) = 140

NEW EQUATION: 7m = 140

DIVIDE 7 to BOTH SIDES

7m/7 = 140/7

CANCEL out: 7/7 because that gives you 1

KEEP: 140/7 because that gives you the value of m

140/7 = m

Solve above and you have the answer of m

140/7 = 20

Therefore, the value of m = 20

Answer: m = 20

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

6 0
3 years ago
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