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Lemur [1.5K]
4 years ago
5

John wants to make cookies. To make cookies, he need 2 over 3 of a cup of flour per batch of cookies. If John has 4 cups of flou

r then how many batch of cookies can John make?
Mathematics
1 answer:
Mnenie [13.5K]4 years ago
6 0
\dfrac{2}{3} \text { cups = } 1 \text { batch}

----------------------------------------------
Find 1 cup :
----------------------------------------------

1 \text { cup = } 1 \div  \dfrac{2}{3} \text { batch}

1 \text { cup = } 1 \times  \dfrac{3}{2} \text { batch}

1 \text { cup = } \dfrac{3}{2} \text { batch}

----------------------------------------------
Find 4 cups :
----------------------------------------------
4 \text { cups = } \dfrac{3}{2} \times 4 \text { batch}

4 \text { cups = } \6 \text { batches}

--------------------------------------------------------------------------------------------
Answer: John can make 6 batches with 4 cups of flour.
--------------------------------------------------------------------------------------------
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Simplify 6/2√3-√6+√6/√3+√2- 4√3/√6-√2
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Answer:

-sqrt(2) + 3 sqrt(3) - sqrt(6)

Step-by-step explanation:

Simplify the following:

(6 sqrt(3))/2 - sqrt(6) + (sqrt(6))/(sqrt(3)) + sqrt(2) - (4 sqrt(3))/(sqrt(6)) - sqrt(2)

6/2 = (2×3)/2 = 3:

3 sqrt(3) - sqrt(6) + (sqrt(6))/(sqrt(3)) + sqrt(2) - (4 sqrt(3))/(sqrt(6)) - sqrt(2)

Rationalize the denominator. (sqrt(6))/(sqrt(3)) = (sqrt(6))/(sqrt(3))×(sqrt(3))/(sqrt(3)) = (sqrt(6) sqrt(3))/3:

3 sqrt(3) - sqrt(6) + (sqrt(6) sqrt(3))/3 + sqrt(2) - (4 sqrt(3))/(sqrt(6)) - sqrt(2)

sqrt(6) sqrt(3) = sqrt(6×3):

3 sqrt(3) - sqrt(6) + (sqrt(6×3))/3 + sqrt(2) - (4 sqrt(3))/(sqrt(6)) - sqrt(2)

6×3 = 18:

3 sqrt(3) - sqrt(6) + (sqrt(18))/3 + sqrt(2) - (4 sqrt(3))/(sqrt(6)) - sqrt(2)

sqrt(18) = sqrt(2×3^2) = 3 sqrt(2):

3 sqrt(3) - sqrt(6) + (3 sqrt(2))/3 + sqrt(2) - (4 sqrt(3))/(sqrt(6)) - sqrt(2)

(3 sqrt(2))/3 = 3/3×sqrt(2) = sqrt(2):

3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) - (4 sqrt(3))/(sqrt(6)) - sqrt(2)

Rationalize the denominator. (-4 sqrt(3))/(sqrt(6)) = (-4 sqrt(3))/(sqrt(6))×(sqrt(6))/(sqrt(6)) = (-4 sqrt(3) sqrt(6))/6:

3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) + (-4 sqrt(3) sqrt(6))/6 - sqrt(2)

The gcd of -4 and 6 is 2, so (-4 sqrt(3) sqrt(6))/6 = ((2 (-2)) sqrt(3) sqrt(6))/(2×3) = 2/2×(-2 sqrt(3) sqrt(6))/3 = (-2 sqrt(3) sqrt(6))/3:

3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) + (-2 sqrt(3) sqrt(6))/3 - sqrt(2)

sqrt(3) sqrt(6) = sqrt(3×6):

3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) - 2/3 sqrt(3×6) - sqrt(2)

3×6 = 18:

3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) - (2 sqrt(18))/3 - sqrt(2)

sqrt(18) = sqrt(2×3^2) = 3 sqrt(2):

3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) - 2/3 3 sqrt(2) - sqrt(2)

Combine powers. (-2×3 sqrt(2))/3 = 2 sqrt(2)×3^(1 - 1) (-1):

3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) - 2 sqrt(2)×3^(1 - 1) - sqrt(2)

1 - 1 = 0:

3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) - 2 sqrt(2)×3^0 - sqrt(2)

3^0 = 1:

3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) - 2 sqrt(2)×1 - sqrt(2)

Add like terms. 3 sqrt(3) - sqrt(6) + sqrt(2) + sqrt(2) - 2 sqrt(2) - sqrt(2) = -sqrt(2) + 3 sqrt(3) - sqrt(6):

Answer: -sqrt(2) + 3 sqrt(3) - sqrt(6)

7 0
3 years ago
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