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3 years ago

Select the choice that translates the following verbal phrase correctly to algebra:

Mathematics

2 answers:

Marat540 [252]3 years ago

5 0

Y - 21
That's the answer.

Rashid [163]3 years ago

5 0

Answer: this is the answer

21 − y

Step-by-step explanation:

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Drag each tile to the correct box. Triangle ABC is inscribed in a circle centered at point o. CK630 B 1180 Order the measures of

Rudik [331]

Answer:

AC < BC < AB

Step-by-step explanation:

Given: BC = 118

AB = 63*2 = 126

AC = 360 - (BC+AB)

AC = 360 - (118+126) = 116

6 0

2 years ago

Find the greatest solution for x+y when x^2+y^2 = 7, x^3+y^3=10

damaskus [11]

Answer:

Step-by-step explanation:

set

$f(x,y)=x+y\\$

constrain:

$g(x,y)=x^2+y^2 = 7\\h(x,y)=x^3+y^3=10$

Partial derivatives:

$f_{x}=1\\f_{y} =1 \\g_{x}=2x \\g_{y}=2y\\h_{x}=3x^2 \\h_{y}=3y^2$

Lagrange multiplier:

$grad(f)=a*grad(g)+b*grad(h)\\$

$\left[\begin{array}{ccc}1\\1\end{array}\right]=a\left[\begin{array}{ccc}2x\\2y\end{array}\right]+b\left[\begin{array}{ccc}3x^2\\3y^2\end{array}\right]$

4 equations:

$1=2ax+3bx^2\\1=2ay+3by^2\\x^2+y^2=7\\x^3+y^3=10$

By solving:

$a=4/9\\b=-2/27\\x+y=4$

Second mathod:

Solve for x^2+y^2 = 7, x^3+y^3=10 first:

$x=\frac{1}{2} -\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} +\frac{\sqrt{13}}{2} \\x=\frac{1}{2} +\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} -\frac{\sqrt{13}}{2} \\x+y=-5\ or\ 1 \or\ 4$

The maximum is 4

6 0

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.

5 0

3 years ago

I need 29 and 30 so please tell me the answer if you know

Nonamiya [84]

Have you tried using the website slader? It has answers for all your questions just search up slader on google then search your book on the website.

7 0

3 years ago

The volume of a cylinder, V, in terms of radius, r,

Igoryamba

Answer:

See explanation

Step-by-step explanation:

Volume of a cylinder = πr²h

Solve for r

V = π * r² * h

r² = V/πh

Find the square root of both sides

r = √V/πh

If height = 6 meters and volume = 300 cubic meters. Find r

r = √V/πh

= √(300/3.14*6)

= √(300/18.84)

= √15.923566878980

r = 3.9904344223381

Approximately,

r = 4 meters

5 0

3 years ago