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gogolik [260]
3 years ago
5

Two friends share 1/6 of a large peach pie. What fraction of the whole pie does each friend get? Let n represent the fraction of

the whole pie each friend gets.
Enter an equation as a multiplication equation.

The equation is n=?

Each friend gets ? of a whole pie.
Mathematics
1 answer:
Firdavs [7]3 years ago
6 0

Answer:

n= 1/12

Step-by-step explanation:

1/6 ÷ 2

1/6×1/2=1/12

Two friends are sharing 1/6 of a pie. That's why it is 1/6 divided by 2. In fractions when you switch from division to multiplication, you switch the denominator and numerator, so 2/1 becomes 1/2. Then, you multiply 1/6 and 1/2. You end up with 1/12, so n= 1/12.

Each friend gets 1/12 of the pie.

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x = 3 or x = 2 or x = -1 or x = -5

Step-by-step explanation:

Solve for x:

((x - 3) (x - 2) (x + 1) (x + 5))/x = 0

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(x - 3) (x - 2) (x + 1) (x + 5) = 0

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x = 3 or x - 2 = 0 or x + 1 = 0 or x + 5 = 0

Add 2 to both sides:

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Subtract 1 from both sides:

x = 3 or x = 2 or x = -1 or x + 5 = 0

Subtract 5 from both sides:

Answer:  x = 3 or x = 2 or x = -1 or x = -5

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f_x = ycos(xy)

f_y = xcos(xy) - ze^{yz}

f_z = -ye^{yz}

we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

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