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aliya0001 [1]
3 years ago
6

Maria will use a 1 fourth-gallon pitcher to fill an empty 2 thirds-gallon jug for her lemonade stand. How much lemonade will she

need in order to completely fill the jug?
A.

between 1 and 2 full pitchers
B.

between 2 and 3 full pitchers
C.

about half of a full pitcher
D.

about 1 fourth of a full pitcher
Mathematics
1 answer:
katen-ka-za [31]3 years ago
6 0

divide 2/3 by 1/4

2/3 / 1/4 = 2/3 * 4/1 = 8/3 = 2 2/3

Answer is B

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Total size of sandwich= L = 4ft
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total no. of servings= 4/0.333 =12 
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4 days ago the gas tank on Ryan's boat was filled with 30 gallons of gasoline today he has 8 gallons of gasoline what is the mea
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4x - 9 = 6 - X<br> What is the answer please
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collect the like terms

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Verify that each equation is an identity (1 - sin^(2)((x)/(2)))/(1+sin^(2)((x)/(2)))= (1+cosx)/(3-cosX)
Allisa [31]

Answer:

Given that we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

By the application of the law of indices and algebraic process of adding a and subtracting a fraction from a whole number, we have;

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

Step-by-step explanation:

An identity is a valid or true equation for all variable values

The given equation is presented as follows;

\dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

From trigonometric identities, we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

\therefore sin^2 \left (\dfrac{x}{2} \right ) = \dfrac{1 - cos (x)}{2}

1 -  sin^2 \left (\dfrac{x}{2} \right ) = 1 - \dfrac{1 - cos (x)}{2} = \dfrac{2 - (1 - cos (x))}{2} = \dfrac{1 + cos (x))}{2}

1 +  sin^2 \left (\dfrac{x}{2} \right ) = 1 + \dfrac{1 - cos (x)}{2} = \dfrac{2 + 1 - cos (x))}{2} = \dfrac{3 - cos (x))}{2}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

3 0
3 years ago
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