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4 years ago

7

I Will pick brainless answer! A recipe for cookies requires 2/3 cup of butter. Rama wants to make 3/4 of the recipe. How many cu

ps of butter should Rama used to make the cookies?

2 answers:

SVEN [57.7K]4 years ago

7 0

Answer:

d: 1/2

Step-by-step explanation:

weeeeeb [17]4 years ago

3 0

Answer:D. 1/2 c

Step-by-step explanation:A recipe is 100% and 3/4 is 75%. 75% of 2/3 is 1/2.

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The length of a rectangle is 25 cm.

likoan [24]

Answer:

20<

Step-by-step explanation:

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4 years ago

sukhopar [10]

Answer:

D. 216 cm

Step-by-step explanation:

1. I assume 14 is a 22 is b and 12 is h

2. Trapezoid Area Formula

$A = \frac{1}{2}(a+b)h$

3. Plug in numbers

$A = \frac{1}{2}(14+22)12$

$216 = \frac{1}{2}(14+22)12$

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3 years ago

Chris types at an average speed of 45 words per minute. He has already typed 1,500 words of his final paper. The paper has to be

lana [24]

Answer:

C. 45x + 1,500 > 3,750

Step-by-step explanation:

You know he has already started on his paper with 1,500 words and he needs to have more words than 3,750 in his final paper which includes the 1,500. Thus, he would have 45x number of words remaining in his paper, according to his speed of 45 words per minute.

Remaining words + 1,500 words must be more than 3,750.

45x + 1,500 > 3,750

The > sign means that something on the left is more than something on the right.

6 0

3 years ago

Read 2 more answers

Acie is filing his federal income tax return with the 1040EZ form, and he

jekas [21]

Answer:4540

Step-by-step explanation:

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4 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.

5 0

3 years ago