All categories

3 years ago

Find the point that splits segment AB in half if point A is located at (4,3) and point B is located at

Mathematics

2 answers:

noname [10]3 years ago

8 0

Answer:

The point that splits the segment AB in half is $C\left ( 1,4.5 \right )$

Step-by-step explanation:

Given: Point A is located at $\left ( 4,3 \right )$ and point $\left ( -2,6 \right )$

To find: Point that splits segment AB in half.

Solution: Let $C\left ( x_{3},y_{3} \right)$ be the point that splits AB in half.

We know that the mid point $\left ( x_{3},y_{3} \right )$ of a line segment joining the points $\left ( x_{1},y_{1} \right )$ and $\left ( x_{2},y_{2} \right )$ is calculated as $\left (\frac{x_{1}+x_{2}}{2},\:\frac{y_{1}+y_{2}}{2} \right )$

Here, $x_{1}=4,\:x_{2}=-2,y_{1}=3,y_{2}=6$

$x_{3}=\frac{4-2}{2}$

$x_{3}=\frac{2}{2}$

$x_{3}=1$

$y_{3}=\frac{3+6}{2}$

$y_{3}=\frac{9}{2}$

$y_{3}=4.5$

Hence, the point that splits the segment AB in half is $C\left ( 1,4.5 \right )$

musickatia [10]3 years ago

5 0

Answer:

(1, 4.5 )

Step-by-step explanation:

The required point is at the midpoint of AB

Use the midpoint formula

Given A(4, 3) and B(- 2, 6 ), then

midpoint = [ 0.5(4 - 2), 0.5(3 + 6) ] = (1, 4.5 )

You might be interested in

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s too easy but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

Differential (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

Differentiation (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

Differential (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$

Now we can substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}$

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

8 0

3 years ago

What is the sum of 1 + 1 = ?

Andrews [41]

The awnswer is 2 yay

3 0

3 years ago

If standard deviation of a given data is 5 and mean is 50 then coefficient of variation (C.V) is equal to *

torisob [31]

Answer:

Step-by-step explanation:

Coefficient of variation is a measure of dispersion, showing the variability of data in relation to the mean.

The Coefficient of variation compares the degree of variation between data points. The coefficient of variation is the ratio of mean to standard deviation. It is given by the formula:

Coefficient of variation = mean / standard deviation

Coefficient of variation = 50 / 5

Coefficient of variation = 10

4 0

3 years ago

Which value of a in the exponential function below would cause the function to stretch?

lesya692 [45]

Answer:

$51251552 | \sqrt[ { {).5644}^{2} }^{25} ]{051} |$