Someone help me find mCD please

List the sides in ΔEFG in order from smallest to largest. Picture includes possible solutions

Juli2301 [7.4K]

The answer for this question is A

8 0

3 years ago

Find the center, vertices, and foci for the ellipse 25x^2 + 64y^2 = 1600

Alisiya [41]

Answer:

The answer to your question is below

Step-by-step explanation:

Data

Equation 25x² + 64y² = 1600

Process

1.- Divide all the equation by 1600

25x²/1600 + 64y²/ 1600 = 1600/1600

-Simplify

x²/64 + y²/ 25 = 1

2.- Equation of a horizontal ellipse

$\frac{x^{2} }{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

3.- Find a, b and c

a² = 64 a = 8

b² = 25 b = 5

-Calculate c with the Pythagorean theorem

a² = b² + c²

-Solve for c

c² = a² - b²

-Substitution

c² = 8² - 5²

-Simplification

c² = 64 - 25

c² = 39

-Result

c = √13

4.- Find the center

C = (0, 0)

5.- Find the vertices

V1 = (-8, 0) V2 = (8, 0)

6.- Find the foci

F1 = (-√13, 0) F2 = (√13, 0)

7 0

3 years ago

the ratio of men to women working for a company is 5 to4 * 225 employees total how many women work for the company

Natali5045456 [20]

63 i believe correct me if i’m wrong somebody. Im possibly WAY off

4 0

3 years ago

Please explain how do you get the answer. Also tell me am I correct or wrong.

Zigmanuir [339]

Your doing everything correctly

3 0

3 years ago

Read 2 more answers

segment BC is an are of a circle with radius 30.0 cm, and point P is at the center of curvature of the arc Segment DA is an arc

Elan Coil [88]

Answer:

The magnetic field is $B = 3.87 *10^{-6} \ T$

Direction is inward

Step-by-step explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The radius of BC is $r_{BC} = 30 \ cm = 0.30 \ m$

The radius of DA is $r _{DA} = 20 \ cm = 0.20 \ m$

The length of CD is $r_{CD} = 10 cm = 0.1 \ m$

The length of AB is $r_{AB} = 10 cm = 0.1 \ m$

The current is $I = 11.4A$

The magnetic field is mathematically represented as

$B = B_{BC} + B_{DA} + B_{CD} + B_{AB}$

$B = \frac{\mu_o I}{4 \pi } [ \frac{\theta_{DA}}{r_{DA}} + \frac{\theta_{BC}}{r_{BC}}+ \frac{\theta_{CD}}{r_{CD}}+\frac{\theta_{AB}}{r_{AB}}]$

Where

$\theta_{BC} = \theta_{DA} = \frac{2\pi}{3}$

Where $\frac{2 \pi}{3} = 120^o$

$\theta_{CD} = \theta_{AB} = 0^o$

so

$B = \frac{\mu_o I}{4 \pi } [ \frac{\frac{2\pi}{3} }{0.20} - \frac{\frac{2\pi}{3}}{0.30}}+ \frac{0}{0.10}+\frac{0}{0.10}]$

$B = 3.87 *10^{-6} \ T$

The direction is into the page

This because the magnitude of the magnetic field due to arc BC whose direction is outward is less than that of DA whose direction is inward

This is because according to Fleming's Left Hand Rule the direction of current is perpendicular to the direction of magnetic field so since current in arc BC and DA are moving in opposite direction their magnetic field will also be moving in opposite direction

3 0

3 years ago