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3 years ago

6

Every 14 days, Debbie’s dog eats 415 pounds of dog food. If Debbie’s dog eats the same amount of food each day, how many pounds

does her dog eat per day?

2 answers:

SIZIF [17.4K]3 years ago

6 0

Answer:

The answer is 7/1 simplified

Step-by-step explanation:

vampirchik [111]3 years ago

4 0

Answer:

i thinks its 29

Step-by-step explanation:

divide 415 by 14

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At a local pizza place, the cost for 3 small pizzas and 3 medium pizzas is $51 The cost for 5 small and 2 medium is $55 How much

faltersainse [42]

Answer:

32

Step-by-step explanation:

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2 years ago

Model the following problem with a quadratic equation. Then solve. Find the length of a side of a square with an area of 75 ft2.

Vaselesa [24]

Answer:

x = 8.6603 m

Step-by-step explanation:

If x is the length of a side of the square, the area of the square will be x^2.

So, if the area of the square is 75 ft2, we can formulate the quadratic equation:

x^2 = 75

Now, solving the equation, we just need to make the square root of 75:

x = sqrt(75) = ±8.6603

x1 = 8.6603

x2 = -8.6603

Now, as x represents the length of a side of the square, and measurements can't be negative, we take only the positive value, so:

x = 8.6603 m

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3 years ago

A car dealership is buying a car for $10,500. They plan to sell it at $15,540. Calculate the percent increase.

lianna [129]

Answer:

67.5 percent (if you're rounding up 68 percent)

Step-by-step explanation:

10,500/15,540 which equals 0.675 turn this into a percent which is 67.5%

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3 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, $\rho_p$

The drag force is given as follows;

$F_D = C_D \times A \times \dfrac{\rho \cdot V^2}{2}$

We have;

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

Therefore;

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

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

$\therefore \dfrac{L_p}{L_m} = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2$

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

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

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

$\dfrac{F_p}{F_m} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3$

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

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

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3 years ago

Line j: y = 1/4x + 4

Sergio039 [100]

A.) Lines J and k because parallel lines have the same slope but NOT y-intercept.

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4 years ago

Read 2 more answers