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Advocard [28]
3 years ago
5

Each of the 27 turtles in the pet store needs to be fed. There is one bag of turtle food that weighs 84 ounces. If each turtle g

ets the same amount of food, how many ounces of turtle food will each turtle get?
Mathematics
2 answers:
Anarel [89]3 years ago
3 0

Step-by-step explanation:

84 ounces / 27 turtles = 3.111 ounces per turtle

Or in fractional form, 3 ¹/₉ ounces per turtle.

sweet [91]3 years ago
3 0

Answer: the answer is 3.1 with the line on top of the 1

Step-by-step explanation:just took the test hope it is help full

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30 students in a class took an algebra test.
lesya [120]

Answer:

70%

Step-by-step explanation:

21/30      p/100

30/30=2,100/30=70

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2 years ago
Is 8/9 bigger than 2/3
o-na [289]

Answer:

Yes!

Step-by-step explanation:

8/9 is equal to .888889 while 2/3 is equal to .66667.

3 0
3 years ago
Read 2 more answers
Simplify the expression completely. 7b+3(4b−2)+6÷3
Karo-lina-s [1.5K]

Answer:

19b-4

Step-by-step explanation:

7b+3(4b-2)+6÷3

Bracket comes first so we distribute 3 over the terms in the bracket

= 7b + 12b -6 +6 ÷ 3

First we solve division

= 7b + 12b -6 +6 ÷ 3

= 7b + 12b -6 +2

= (7b+12b)+(-6+2)

= 19b -4

This can't be simplified further as it can't be solved anymore

4 0
3 years ago
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NO LINKS <br> Please help
Triss [41]

Answer:

V≈1287.88

A≈1152.98

Step-by-step explanation:

7 0
2 years ago
Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete
Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. 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}}

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.)

__

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}

__

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

__

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}

__

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}}

__

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}

3 0
3 years ago
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