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Digiron [165]
3 years ago
8

Deanna uses only fruit juice and sparkling water to make punch for a school dance. She uses 6 liters of fruit juice. She uses 0.

6 as much sparkling water as fruit juice.
How much fruit punch does she make?
Mathematics
2 answers:
Softa [21]3 years ago
3 0

Answer: 3.6 liters

Step-by-step explanation:

You know that:

- She uses 6 liters of fruit juice.

- She uses 0.6 as much sparkling water as fruit juice.

Therefore, let's call x the amount of fruit punch she made.

Keeping the above on mind, you can solve the problem by multiplying the 6 liter of fruit juice she uses by 0.6.

Therefore the result is:

x=(6liter)(0.6)\\x=3.6liters

Sloan [31]3 years ago
3 0

<u>Answer:</u>

Deanna makes 9.6 liters of fruit punch.

<u>Step-by-step explanation:</u>

We are given that Deanna uses 6 liters of fruit juice and 0.6 as much sparkling water as fruit juice.

We are to find out the amount of fruit punch she makes.

Liters of fruit juice used = 6

Liters of sparkling water used = 0.6 * 6 = 3.6

Liters of fruit punch made = 6 + 3.6 = 9.6 liters

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Answer:

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Step-by-step explanation:

f(2)=3(2)-1

f(2)=6-1

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We then get:

g(x)=f(2)

subsitute 5 for f(2)

g(x)=5

now solve

5=2x-3

add 3

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g(4)=f(2)

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3 years ago
Which number added to -14 will equal 9?​
Vlada [557]

Answer:

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2 years ago
A 1/17th scale model of a new hybrid car is tested in a wind tunnel at the same Reynolds number as that of the full-scale protot
Olegator [25]

Answer:

The ratio of the drag coefficients \dfrac{F_m}{F_p} is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, c_{m} = The drag coefficient of the prototype, c_{p}

The medium of the test for the model, \rho_m = The medium of the test for the prototype, \rho_p

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We have;

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Therefore;

\dfrac{L_p}{L_m}  = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2

\dfrac{L_p}{L_m}  =\dfrac{17}{1}

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\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2

\dfrac{F_p}{F_m}  = \dfrac{c_p \times A_p \times  \dfrac{\rho_p \cdot V_p^2}{2}}{c_m \times A_m \times  \dfrac{\rho_m \cdot V_m^2}{2}} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}

\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2

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\dfrac{F_m}{F_p}  = \left( \left\dfrac{1}{17} \right)^3= (1/17)^3 ≈ 0.0002

The ratio of the drag coefficients \dfrac{F_m}{F_p} ≈ 0.0002.

5 0
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frutty [35]
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------------------
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5y = 2.5
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ollegr [7]

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