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3 years ago

Simplify the complex fraction: ((3x-7)/x^2)/(x^2/2)+(2/x)

1 answer:

6 0

Simplify the complex fraction: ((3x-7)/x^2)/(x^2/2)+(2/x)

$\dfrac{ \dfrac{3x-7}{x^2} }{ \dfrac{x^2}{2} } + \dfrac{2}{x} =\qquad \qquad x \neq 0\qquad x^4 \neq 0 \\ \\ \\ \dfrac{(3x-7)*2}{x^2*x^2}+ \dfrac{2}{x} = \\ \\ \\ \dfrac{6x-14}{x^4}+ \dfrac{2}{x} =\\ \\ \\ \dfrac{ 6x-14 }{ x^4 }+ \dfrac{2(x^3)}{x(x^3)}=\\ \\ \\ \boxed{ \dfrac{ 6x-14 +2x^3}{ x^4 } }$

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Please help me with this!!!!!

vekshin1

Answer:

y = 4

Step-by-step explanation:

Note that A and B have the same x-coordinate: 1. This means that AB is a vertical line through x = 1. The perpendicular bisector of AB must be a horizontal line situated halfway between y = 3 and y = 5. Thus, it's just y = 4.

Review: y = 4 bisects AB and represents a horizontal line.

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3 years ago

If x - 5y = 6, what is the value of y when X= -4? A:-2 B-2/5 C:2/5 D:2

zysi [14]

Answer:

I think the answer is D. 2

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2 years ago

Read 2 more answers

PLEASE HELP

Oksana_A [137]

Answer:

A racer has just completed his first lap at a speed of 198m/s. Now his speed decrease or he decelerateted at 195m/s after 2s of his completion of first lap, as soon as he saw a curve path(line segment R). He then moved at a constant speed of 195m/s for the next 2s(line segment s). Then the race track after 4s of the completion of first lap was straight, so the racer accelerated to 200m/s within 2s( line segment T) and traveled at a constant speed for the next 6s( line segment U). After 12s from the first lap completion, he immediately saw a curve track, so he decelerateted to the speed of 194m/s within 3s( line segment V).

Hope it helped.

please mark brainliest

4 0

2 years ago

Urgent. There are 12 coloured counters in a bag. The counters are black, white or grey A counter is chosen at random. The probab

Nastasia [14]

Answer:

1/12

Step-by-step explanation:

Needed information

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

The sum of the probabilities of all outcomes must equal 1

Solution

We are told that the probability that the counter is not black is 3/4.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is black by subtracting 3/4 from 1:

$\sf P(counter\:not\:black)=\dfrac34$

$\implies \sf P(counter\:black)=1-\dfrac34=\dfrac14$

We are told that the probability that the counter is not white is 2/3.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is white by subtracting 2/3 from 1:

$\sf P(counter\:not\:white)=\dfrac23$

$\implies \sf P(counter\:white)=1-\dfrac23=\dfrac13$

We are told that there are black, white and grey counters in the bag. We also know that the sum of the probabilities of all outcomes must equal 1. Therefore, we can work out the probability the counter is grey by subtracting the probability the counter is black and the probability the counter is white from 1:

$\begin{aligned}\sf P(counter\:grey) & = \sf1-P(counter\:black)-P(counter\:white)\\\\ & =\sf 1-\dfrac14-\dfrac13\\\\ & = \sf \dfrac{12}{12}-\dfrac{3}{12}-\dfrac{4}{12}\\\\ & = \sf \dfrac{12-3-4}{12}\\\\ & = \sf \dfrac{5}{12}\end{aligned}$

7 0

2 years ago

Two teachers compared the mean absolute deviation of the scores from the semester exam they gave to their students. Mrs. Alvarez

Monica [59]

Answer:

C. Mrs. Alvarez's scores were less spread out than Mr. Crawford's scores.

Step-by-step explanation:

Mean Absolute Deviation is one of the Statistical measures which we can you to determine the variation that exist amongst a given set of data

Mean Absolute Deviation can be defined as how far or the distance between one set of data to another set of data.

The smaller the Mean Standard Deviation, the lower the degree of variation in the set of data. The data is less spread out

The larger the Mean Standard Deviation, the higher the degree of variation in the set of data. The data is Largely spread out

We are told in the question that:

Mrs. Alvarez's scores had a lower mean absolute deviation than Mr. Crawford's scores. Our conclusion would be that Mrs. Alvarez's scores were less spread out than Mr. Crawford's scores.

Option 2 is correct.

3 0

3 years ago