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valina [46]
3 years ago
10

PLEASE ANYONE I NEED YOUR HELP. For the points A(-2, 10) and B(-4,6). Find each of the following.

Mathematics
2 answers:
Rashid [163]3 years ago
8 0

Answer:

a. _ √20 , about 4.472136

b - (-3, 8)

c- Slope of 2

Step-by-step explanation:

Calculator

almond37 [142]3 years ago
5 0

Answer:

a. AB=2\sqrt{5}

b. (-3,8)

c. 2

Step-by-step explanation:

You have the points:

A(-2,10)

where i will call: x_{1}=-2 and y_{1}=10

B(-4,6)

where i will call: x_{2}=-4 and y_{2}=6

for our calculations we are going to need the distance in x between the points (\Delta x )and the distance in y between the points (\Delta y):

\Delta x =|x_{2}-x_{1}|=|-4-(-2)|=|-4+2|=|-2|=2\\\Delta y =|y_{2}-y_{1}|=|6-10|=|-4|=4

a. To find AB (the distance between point A and point B) you need The Pythagorean Theorem:

(AB)^2=(\Delta x)^2+(\Delta y)^2\\(AB)^2=(2)^2+(4)^2\\(AB)^2=4+16\\\\AB=\sqrt{20}\\ AB=2\sqrt{5}

b. to find the coordinates of the midpoint we average the x-coordinates and the y coordinates

x_{mid}=\frac{x_{1}+x_{2}}{2} =\frac{-2-4}{2}=\frac{-6}{2} =-3\\y_{mid}=\frac{y_{1}+y_{2}}{2} =\frac{10+6}{2}=\frac{16}{2} =8\\

so the midpoint (x_{mid},y_{mid}) is at: (-3,8)

c. For the slope we use the slope formula:

slope=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{6-10}{-4-(-2)}=\frac{-4}{-2}=2

The slope is equal to 2.

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3 years ago
Hi, umm pls help<br><br> 1/6(x-5)=1/2(x+6), solve for x <br><br> And can you please explain
Mumz [18]

Answer:

By solving the equation \frac{1}{6}(x-5)=\frac{1}{2}(x+6) we found that   x=\frac{23}{2} \text { or }=11 \frac{1}{2}

<u>Explanation:  </u>

Given equation,

\frac{1}{6}(x-5)=\frac{1}{2}(x+6) to find x  

Step: 1 Cross multiply the denominators 2(x-5)=6(x+6)  

Step: 2 Open brackets and simplify the term 2x-10=6x+36  

Step: 3 Bring x terms on one side and simplify the equation the equation obtained is following,  

2x-6x=36+10

 -4x=46  

-x=\frac{46}{4}=\frac{23}{2}=11 \frac{1}{2}

Therefore, we solved x value from the given equation \frac{1}{6}(x-5)=\frac{1}{2}(x+6) as x=\frac{23}{2} \text { or }=11 \frac{1}{2}.

4 0
3 years ago
Describe a scenario where the function value exists at x=c, but the limit does not exist.
Vikki [24]

The scenario can be described using a piecewise function like:

f(x) = 1/x                  if  x < c.

f(x) = x                    if   x = c

f(x) = 1/(x + 73)       if x > c.

<h3>When the value exists but the limit does not?</h3>

Remember that the limit only exists if the limit from left and the limit from the right give the same value.

Then, we can just define a piecewise function of the form:

f(x) = 1/x                  if  x < c.

f(x) = x                    if   x = c

f(x) = 1/(x + 73)       if x > c.

Clearly, this is not a continuous function.

Notice that:

f(c) = c.\\\\ \lim_{x \to c^{-}} f(x) = 1/c\\\\ \lim_{x \to c^{+}} f(x) = 1/(c + 73)

So the limits from left and right are different, then:

\lim_{x \to c^{}} f(x)

Does not exist.

If you want to learn more about limits:

brainly.com/question/5313449

#SPJ1

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