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2 years ago

Math Help Part 2 (Please Help)

Mathematics

1 answer:

GenaCL600 [577]2 years ago

7 0

Answer:

$AB = 104$

$BE = 48$

$CG = 60$

Step-by-step explanation:

see attachment for step-by-step

(Brainly won't let me post as it thinks there are inappropriate words!)

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Help me with this, please!!!!

user100 [1]

Answer: D (1.5,0.5)

Step-by-step explanation:

It's not negative so cross out all those questions that leaves you with C and D. Now we can see that it's past the 1 point so it's in the 1.0 range and0. Range and Cs answer has 3 y on it which is incorrect. So that leaves you with D.

7 0

2 years ago

mr hill has 27 students in his class and mr chang has 24 students in his class.both classes will be divided into equal sized tea

Svetlanka [38]

Answer:

mr hill: 9

mr chang: 12

Step-by-step explanation:

mr hill can have 3 teams of 9 and

mr chang can have 2 teams of 12

5 0

3 years ago

Read 2 more answers

1. Write the equation of the line passing through the point (-2, 2) that is

Artyom0805 [142]

creo que quieres decir la suma

si es la suma eso da 2.8

5 0

2 years ago

PLEASE HELP Which of the following are true if event A and B are independent

makkiz [27]

Answer:

a) $P(B/A) = P(B)$

Step-by-step explanation:

Step(i):-

Given that A and B are independent events

P(A∩B) =P(A) P(B)

we will use the conditional probability

$P(A/B) = \frac{P(AnB)}{P(B)}$

$P(B/A) = \frac{P(AnB)}{P(A)}$

Step(ii):-

Given that A and B are independent events

P(A∩B) =P(A) P(B)

$P(B/A) = \frac{P(AnB)}{P(A)}$

$P(B/A) = \frac{P(A)P(B)}{P(A)} = P(B)$

$P(B/A) = P(B)$

3 0

2 years ago

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = xyi + 5zj + 7yk

REY [17]

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

where $S$ is the surface with $C$ as its boundary. The curl is

$\nabla\times\vec F(x,y,z)=2\,\vec\imath-x\,\vec k$

Parameterize $S$ by

$\vec\sigma(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(8-u\cos v)\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Then take the normal vector to $S$ to be

$\vec\sigma_u\times\vec\sigma_v=u\,\vec\imath+u\,\vec k$

Then the line integral is equal to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9(2\,\vec\imath-u\cos v\,\vec k)\cdot(u\,\vec\imath+u\,\vec k)\,\mathrm du\,\mathrm dv$

$\displaystyle=\int_0^{2\pi}\int_0^9(2u-u^2\cos v)\,\mathrm du\,\mathrm dv=\boxed{162\pi}$

7 0

2 years ago