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avanturin [10]
3 years ago
8

At camp, ¼ of the children are divided into 3 equal groups for crafts. What fraction of the children at camp are in EACH group?

Mathematics
2 answers:
zloy xaker [14]3 years ago
7 0

Answer:

3/4?

Step-by-step explanation:

maybe not but i think this is it

erma4kov [3.2K]3 years ago
5 0
3/4........…............
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A fish tank has black and white fish in the ratio of 2 to 6 as shown below Which is another ratio that can describe the black to
ira [324]
1:3, Because you can simplify the numbers by dividing by two.
 
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3 years ago
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11111nata11111 [884]

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Step-by-step explanation:

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3 years ago
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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}}

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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}

3 0
3 years ago
Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.
maks197457 [2]
F(g(x)) = [(-7x-8)/(x-1) - 8} / [(-7x - 8)/(x-1) + 7] =

[(-7x - 8 - 8(x-1)) / (x-1)] / [(-7x - 8 + 7(x-1)) / (x-1)] = (-15x) / (-15) = x.

g(f(x)) = [-7*(x-8)/(x+7) - 8] / [(x-8)/(x+7) - 1] =

[(-7x + 56 -8*(x+7)) / (x+7)] / [(x - 8 - (x + 7)) / (x+7)] = (-15x) / (-15) = x.

So since f(g(x)) = g(f(x)) = x we can conclude that f and g are inverses.
5 0
3 years ago
Which expression has the same value as the one below?
sasho [114]
The answer is C. 33-10
7 0
3 years ago
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