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2 years ago

Please help! is there any solutions?

Mathematics

1 answer:

Digiron [165]2 years ago

5 0

The equations will only have one solution because they have different slopes and also next time use desmos for those kinds of problems it’ll always help

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What is the equation for the locus of points 10 units from the origin?

Lesechka [4]

Answer:

d or b no a tho

Step-by-step explanation:

8 0

3 years ago

How to find the x value given the y value on a ti 84?

OlgaM077 [116]

X = 9
y = x²
y=81
=
$\sqrt{81}$
=2

5 0

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, $\rho_p$

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

2 years ago

Evaluate V= bh divided by 3 for B= 9 in 2 and h=32 in

Luba_88 [7]

It would be 96 because V=(9)(32)= 288/3=96

8 0

3 years ago

A sign on the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 10 cars needs to

Archy [21]

Answer:

0.2916, 0.1488, 0.0319

Step-by-step explanation:

Given that a sign on the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 10 cars needs to have oil added.

Since each trial is independent there is a constant probability for any random car to need oil is 0.10

Let X be the number of cars that need oil

A) Here X is BIN(4,0.1)

$P(X=1) = 4C1(0.1)(0.9)^3 \\= 0.2916$

B) Here X is Bin (8, 0.1)

$P(x=2) = 8C2 (0.1)^2 (0.9)^6\\\\=0.1488$

C) Here X is Bin (20,5)

$P(x=5) = 20C5 (0.1)^5 (0.9)^{15} \\=0.0319$

3 0

2 years ago