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Bogdan [553]
2 years ago
12

I NEED THE ANSWR ASAP: Write the number as a power with a base of 5. 125^3

Mathematics
1 answer:
EastWind [94]2 years ago
5 0

Answer:

5^9

Step-by-step explanation:

I had this question before.

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Area of a circular ring: A = 4<img src="https://tex.z-dn.net/?f=%20%5Cpi%20" id="TexFormula1" title=" \pi " alt=" \pi " align="a
OleMash [197]
1a)      A = 4πpw
      /4πw = /4πw
  A / 4πw = p

1b) A = 4πpw
     22 = 4πp(2)
       p = 11/4π (≈0.87)

2a)        P = 2πr + 2x
      P - 2x = 2πr
        /2π      /2π
 P-2x / 2π = r

2b)    P = 2πr + 2x
      440 = 2πr + 2(110)
          r = 110/π (≈35.014)
      

      
8 0
3 years ago
The average number of points a basketball team scored for
Ganezh [65]
Average is

(2x + X+6)/3=63
(2x+x+6)= 189
3x+6=189
3x=183 61

X=61
Scores for the game are: 61, 61, and 67
3 0
3 years ago
How many terms are in
tatyana61 [14]

Answer:

4

Step-by-step explanation:

4 0
3 years ago
HELP ASAP
galben [10]

I don’t have a very detailed explanation, however the answer should be “A 90-degree counterclockwise rotation about the origin followed by a translation 1 unit to the left”.

You can clearly see a 90° angle here, it’s obvious. Also, the polygon switch counterclockwise. so, you can see.

hope I helped ^^

3 0
3 years ago
Read 2 more answers
Businesses deposit large sums of money into bank accountsImagine an account with $10 million dollars in it.
adoni [48]

again, let's assume daily compounding means 365 days per year.

~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ t=years\dotfill &1 \end{cases} \\\\\\ I = (10000000)(0.0212)(1)\implies \boxed{I=212000}

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\dotfill &365\\ t=years\dotfill &1 \end{cases}

A=10000000\left(1+\frac{0.0212}{365}\right)^{365\cdot 1}\implies A\approx 10214256.88 \\\\\\ \underset{\textit{earned interest amount}}{10214256.88~~ - ~~10000000 ~~ \approx ~~ \boxed{214256.88}}

what's their difference?  well

\stackrel{\textit{compounded daily}}{\approx 214256.88}~~ - ~~\stackrel{\textit{simple interest}}{212000}\implies \boxed{2256.88}

7 0
2 years ago
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