Mathematics, 10.10.2021 20:10 haileysolis5

What is 6xy . 12 over x 100 over 6

Answers: 2

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Answers

Simplifying
X2 + -6xy + -12 = 0

Reorder the terms:
-12 + X2 + -6xy = 0

Solving
-12 + X2 + -6xy = 0

Solving for variable 'X'.

Move all terms containing X to the left, all other terms to the right.

Add '12' to each side of the equation.
-12 + X2 + 12 + -6xy = 0 + 12

Reorder the terms:
-12 + 12 + X2 + -6xy = 0 + 12

Combine like terms: -12 + 12 = 0
0 + X2 + -6xy = 0 + 12
X2 + -6xy = 0 + 12

Combine like terms: 0 + 12 = 12
X2 + -6xy = 12

Add '6xy' to each side of the equation.
X2 + -6xy + 6xy = 12 + 6xy

Combine like terms: -6xy + 6xy = 0
X2 + 0 = 12 + 6xy
X2 = 12 + 6xy

Simplifying
X2 = 12 + 6xy

Reorder the terms:
-12 + X2 + -6xy = 12 + 6xy + -12 + -6xy

Reorder the terms:
-12 + X2 + -6xy = 12 + -12 + 6xy + -6xy

Combine like terms: 12 + -12 = 0
-12 + X2 + -6xy = 0 + 6xy + -6xy
-12 + X2 + -6xy = 6xy + -6xy

Combine like terms: 6xy + -6xy = 0
-12 + X2 + -6xy = 0

The solution to this equation could not be determined.

Answer from: Quest

a) one student is chosen from classroom a, then a student is chosen from classroom b.

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b. this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes. this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes. this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes. this makes them dependent events.

Answer from: Quest

answer: the answer is (b) amie should have multiplied

$54~\textup{m}^3~\textup{by}~\dfrac{4}{3}\pi.$

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

$v_s=\pi r^2h=\pi r^3,$

and the volume of the cylinder will be

$v_c=\dfrac{4}{3}\pi r^3=54~\textup{m}^3.$

therefore, we have

$v_c=\dfrac{4}{3}v_s=\dfrac{4}{3}\times 54=72~\textup{m}^3.$

thus, ammie should have multiplied the volume of the sphere $54~\textup{m}^3~\textup{by}~\dfrac{4}{3}\pi.$

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