Answer:
more than 20 dollars
Step-by-step explanation:
It doesn't matter how much money was in the account before; it matters that the account was overdrawn, which means whatever amount of money was taken out made the balance go negative. Then, the problem says that after the deposit was made, the account had a balance of 20 dollars, which means the person must have put in at least 20 dollars to make the account go from negative to zero, and then to 20.
The answer is a. 5 x 10³.
3.75 x 10^5 = 375,00.( population of Greenville)
375,000 ÷ 75 = 5,000 ( population of Fairview)
5 x 10³(1,000)= 5,000.
Answer:
Not a full question
Step-by-step explanation:
Where’s the graph?
Answer:
- b/a
- 16a²b²
- n¹⁰/(16m⁶)
- y⁸/x¹⁰
- m⁷n³n/m
Step-by-step explanation:
These problems make use of three rules of exponents:
![a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}](https://tex.z-dn.net/?f=a%5Eba%5Ec%3Da%5E%7Bb%2Bc%7D%5C%5C%5C%5C%28a%5Eb%29%5Ec%3Da%5E%7Bbc%7D%5C%5C%5C%5Ca%5E%7B-b%7D%3D%5Cdfrac%7B1%7D%7Ba%5Eb%7D%20%5Cquad%5Ctext%7Bor%7D%20%5Cquad%20a%5Eb%3D%5Cdfrac%7B1%7D%7Ba%5E%7B-b%7D%7D)
In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)
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1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.
![\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}](https://tex.z-dn.net/?f=%5Cdfrac%7Bb%5E%7B-2%7D%7D%7Bab%5E%7B-3%7D%7D%3D%5Cdfrac%7Bb%5E%7B-2-%28-3%29%7D%7D%7Ba%7D%3D%5Cdfrac%7Bb%7D%7Ba%7D)
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2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.
![\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B1%7D%7B4ab%7D%5Cright%29%5E%7B-2%7D%3D%5Cdfrac%7B1%7D%7B4%5E%7B-2%7Da%5E%7B-2%7Db%5E%7B-2%7D%7D%3D16a%5E2b%5E2)
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3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.
![\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B4mn%7D%7Bm%5E%7B-2%7Dn%5E6%7D%5Cright%29%5E%7B-2%7D%3D%5Cleft%284m%5E%7B1-%28-2%29%7Dn%5E%7B1-6%7D%7D%5Cright%29%5E%7B-2%7D%3D%5Cleft%284m%5E3n%5E%7B-5%7D%7D%5Cright%29%5E%7B-2%7D%5C%5C%5C%5C%3D4%5E%7B-2%7Dm%5E%7B-6%7Dn%5E%7B10%7D%3D%5Cdfrac%7Bn%5E%7B10%7D%7D%7B16m%5E6%7D)
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4. This works the same way the previous problem does.
![\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7Bx%5E%7B-4%7Dy%7D%7Bx%5E%7B-9%7Dy%5E5%7D%5Cright%29%5E%7B-2%7D%3D%5Cleft%28x%5E%7B-4-%28-9%29%7Dy%5E%7B1-5%7D%5Cright%29%5E%7B-2%7D%3D%5Cleft%28x%5E%7B5%7Dy%5E%7B-4%7D%5Cright%29%5E%7B-2%7D%5C%5C%5C%5C%3Dx%5E%7B-10%7Dy%5E%7B8%7D%3D%5Cdfrac%7By%5E8%7D%7Bx%5E%7B10%7D%7D)
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5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.
![\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}](https://tex.z-dn.net/?f=%5Cdfrac%7Bm%5E7n%5E3%7D%7Bmn%5E%7B-1%7D%7D%3D%5Cdfrac%7Bm%5E7n%5E3n%7D%7Bm%7D)