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kirill [66]
4 years ago
6

83.9−0.584, please help me with this problem

Mathematics
2 answers:
g100num [7]4 years ago
7 0

Answer:

Step-by-step explanation:

83.900-

  0.584

-------------

 83.316

Serggg [28]4 years ago
5 0

Answer:

83.9-0.584=83.316 is the answer

Step-by-step explanation:

i hope this will help you :)

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Kendra has $ 4.85 in nickels and quarters. If she has 5 more quarters than nickels, how many of each coin does she have?
yarga [219]

If she were to add 5 nickels ($0.25), then she would have an equal number of nickels and quarters, and her total would be $5.10. An equal number of nickels and quarters gives a multiple of $0.30. $5.10 is 17 times $0.30. The number of quarters Kendra has must be 17, but she has 5 fewer nickels, so

Kendra has 17 quarters and 12 nickels.

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3 years ago
What would be the result of 100 plus 4% for 12 years
VashaNatasha [74]

Answer:

100.04

Step-by-step explanation:

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Use the point and the slope to graph each line. Write the equation of the line. The line contains point (2,-2) and is perpendicu
lisabon 2012 [21]

Answer:

x + y = 0

Step-by-step explanation:

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4 0
3 years ago
$20000 is invested in an account that earned 6% p.A. Compounding yearly for 3 years. The interest rate then went up to 8% p.A. F
GuDViN [60]

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}

\bf A=20000\left(1+\frac{0.06}{1}\right)^{1\cdot 3}\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}

\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}

\bf A=2382.032\left(1+\frac{0.08}{1}\right)^{1\cdot 4}\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762

4 0
3 years ago
Need help with math's home work
Aleonysh [2.5K]
1. The best thing to do is to find out what 1% is. 235/100= 2.35
As you're looking for 27%, you then multiply 2.35*27= 63.45
As you're increasing it, you've then got to add this to the original amount.
235+63.45= 298.45

2. Again, find 1%. 24/100= 0.24
0.24*9= 2.16
24+2.16= 26.16

3. Find 1%. 1120/100= 11.2
As you're looking for 13.5%, you then multiply this by 13.5
11.2*13.5= 151.20
As you're subtracting, you then take this away from the original number.
1120-151.20= 968.80

4. This time, you can find 10%.
0.057/10= 0.0057
0.0057*5.5= 0.03135.
You've then got to subtract this as you're decreasing the number.
0.057-0.03135= 0.02565.

Sally's investment is now worth £585. (Find the 30% and add it on) and Susie's investment is now worth £585 (Find the 10% and subtract). They both have the same amount of money.

Hope this helps :) 
8 0
3 years ago
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