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3 years ago

What is the slope line that would be parallel to the given line???

Mathematics

1 answer:

Serjik [45]3 years ago

6 0

Slope of a line parallel to given line is equal to slope of given line.

<u>SOLUTION:</u>

Given that, we have to find the slope of any line that is parallel to a line in the co – ordinate plane.

We know that, slope of a line is tan( angle of a line made between that line and the x – axis)

So, consider two lines which are parallel , then, angles made by both the parallel lines will be equal with the x – axis.

Then, tan( angle by 1st line) = tan( angle by 2nd line)

Which means that, slope of two lines will be equal.

Hence, slope of a line parallel to given is equal to slope of given line.

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4. Convert 23 miles per hour to kilometers per minute. (1 mile = 1.609 kilometers)

Karo-lina-s [1.5K]

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3 years ago

Type the correct answer in each box. Use sigma notation to represent the following series for the first 10 terms. 3 + (-6) + 12

klemol [59]

Answer:

Kindly check attached picture

Step-by-step explanation:

Given the following :

First term (a) = 3

Second term = - 6

third term = 12

Fourth term = - 24

We can obtain the common ratio of the series using the relation :

(2nd term/ 1st term) = (3rd term/2nd term) = (4th term/ 3rd term) =....

(-6/3) = (12/-6) = (-24/12)

-2 = - 2 = - 2

Further explanation can be found in attached picture

7 0

3 years ago

Find the value of each trigonometric ratio. Simplify

AysviL [449]

Tan = opp/adj

Tan A = 16/12

7 0

2 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Mkey [24]

For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .