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1 year ago

Consider the following points on cartesian plane. A (1;2) BC3; 1) and CG-3; 4) prove that Points A, B and C are collinear.

Mathematics

1 answer:

Arlecino [84]1 year ago

5 0

Step-by-step explanation:

Collinear points have the same gradient

find the gradient(slope) between AB first

$= \frac{1 - 2}{3 - 1} \\ = \frac{ - 1}{2}$

Find the one for BC

$= \frac{4 - 1}{ - 3 - 3} \\ = \frac{3}{ - 6} \\ = \frac{ - 1}{2}$

Find the gradient between AC now

$= \frac{4 - 2}{ - 3 - 1} \\ = \frac{2}{ - 4} \\ = \frac{ - 1}{2}$

YOUR LAST STATEMENT NOW WILL BE

POINTS A,B AND C ARE COLLINEAR SINCE THE GRADIENTS BETWEEN THEM IS THE SAME!!!

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A chemist examines 15 geological samples for potassium chloride concentration. The mean potassium chloride concentration for the

NikAS [45]

Answer:

$0.376-2.145\frac{0.0012}{\sqrt{15}}=0.375$

$0.376+2.145\frac{0.0012}{\sqrt{15}}=0.377$

And we are 95% confident that the true mean of chloride concentration is between 0.375 and 0.377 cc/ cubic meter

Step-by-step explanation:

Data provided

$\bar X=0.376$ represent the sample mean for the chloride concentration

$\mu$ population mean (variable of interest)

s=0.0012 represent the sample standard deviation

n=15 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since we need to find the critical value first we need to begin finding the degreed of freedom

$df=n-1=15-1=14$

The Confidence is 0.95 or 95%, the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , and the critical value for this case with a t distribution with 14 degrees of freedom is $t_{\alpha/2}=2.145$

And the confidence interval is:

$0.376-2.145\frac{0.0012}{\sqrt{15}}=0.375$

$0.376+2.145\frac{0.0012}{\sqrt{15}}=0.377$

And we are 95% confident that the true mean of chloride concentration is between 0.375 and 0.377 cc/ cubic meter

7 0

3 years ago

On the first day it was posted online, a music video got 1750 views. The

dimulka [17.4K]

The answer is actually 2187.36. So, rounded, it is 2187

6 0

2 years ago

Read 2 more answers

a library has an average of 510 visitors on Sundays and 240 visitors on other days. The average number of visitors per day in a

serious [3.7K]

Answer:

285

Step-by-step explanation:

personally this is how i would do this question but there are other ways.

i would start off by seeing how many sundays there are in the month (which is 5) then the number of other days which is 25. i'd then multiply 5 by 510 which is 2550. add that to 240 times 25 (which is 6000) and all of that together is 8550. divide that by the number of the days in the month (30) and then you'll get the daily avergae for the month

hope this helped:)

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2 years ago

Factoring trinomials

Tcecarenko [31]

$c^2$ $2c^2 + 13c +21 \\4c^2 + 13(2)c + 42\\(2c + 6)(2c+7)\\(c+3)(2c+7)$

This is a way to factoring trinomials (there exist different equivalent methods).

Multiply the trinomial but the term accompanying $c^2$ . This is the second line. Then, you could take the square of the $4c^2$ , ant try to create a factor () () that will correspond to the expression in the second line. That is, we want $4c^2 + 13(2) c + 42 = (2c + ?) (2c + ?)$

In ? we put the corresponding numbers that, if we multiply them we will obtain 42, and if we add them we will obtain 13. This numbers are 6 and 7. Then, we have $(2c + 6) (2c +7)$

The last step is divide by the number that we multipy in the first step.

4 0

3 years ago

Can somebody please solve this for me and show the work? it doesn't look too hard I'm just dumb.

lana [24]

Answer:

x = 7

Step-by-step explanation:

vertical angles are pairs of opposite angles made by intersecting lines. and they are congruent

x + 16 = 4x - 5

x - 4x = - 5 - 16

- 3x = - 21

- x = - 21/3

- x = - 7

x = 7

hope it helps

and please just focus on your questions and use knowledge

8 0

3 years ago

Consider the following points on cartesian plane. A (1;2) BC3; 1) and CG-3; 4) prove that Points A, B and C are collinear. ​

Consider the following points on cartesian plane. A (1;2) BC3; 1) and CG-3; 4) prove that Points A, B and C are collinear.