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shutvik [7]
3 years ago
15

PLZZZ HELP!! DESPRATE!!! 15PTS!!!!!

Mathematics
2 answers:
s344n2d4d5 [400]3 years ago
7 0

The correct answer is C, x is greater than -5. In order to solve this, you have to solve for x by simplifying both sides of the equation. Then you have to isolate the variable. Hope this helped!

alekssr [168]3 years ago
5 0

Answer:

x > -5

Step-by-step explanation:

Let's begin by combining like terms.  Adding 2x to both sides of this inequality yields -4 < 5x + 21.  Next we combine the constant terms, obtaining -25 < 5x, or x > -5.  This correspondes to answer choice C.

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Sharon is decreasing the size of a diagram of a leaf that is 30 centimeters long by 10 centimeters wide. If the reduced diagram
ivanzaharov [21]
10 : 30
4: x [x is the length]
10/4=2.5
30/2.5=x=12
length of the diagram is 12cm

5 0
3 years ago
Which answer is an equation in point-slope form for the given point and slope? Point: (1,9); Slope:5
JulijaS [17]
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5 0
3 years ago
Read 2 more answers
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
What problem makes 49
torisob [31]
45 + 4 makes 49 if that counts
6 0
3 years ago
Someone help me with this please
storchak [24]

Answer:

x = 114

Step-by-step explanation:

180 - 105 = 75

39 + 75 = x

x = 114

8 0
2 years ago
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