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3 years ago

What are the coordinates of the point (−1, 6) after a reflection across the x-axis?

Mathematics

2 answers:

Bess [88]3 years ago

5 0

The reflection of a point A(a, b) across the x-axis, is the point A'(a, -b).

Thus the point P(-1, 6) is reflected to P'(-1, -6).

Check the picture.

marissa [1.9K]3 years ago

5 0

The coordinates of the point (−1, 6) after a reflection across the x-axis is (−1, −6)

<h3>Further explanation </h3>

A reflection is a common type of transformation. To reflect a point across the x-axis, we consider this axis to be a mirror.

The x-axis runs left to right whereas the y-axis runs up and down. If you want to reflect some function, just picture it being flipped over. From (-1,6) which is located in the upper left and if we reflect it over the x-axis, both points are now in the bottom left quadrant where everything is negative. If we reflect a point across the x-axis, the x-coordinate remains same, whereas the y-coordinate changes into its opposite. So if we have a point , the reflection of this point across the x-axis is the point . Finally, the point (−1, 6) changes into (−1, −6).

When we reflect a point across the x-axis, the x-coordinate remains the same, whereas the y-coordinate is transformed into its opposite, the sign is changed. The simple way if we forget the rules for reflections when graphing, fold your paper along the x-axis to see where the new figure will be located. The reflection of the point (x,y) across the x-axis is the point (x,-y).

<h3>Learn more</h3>

Learn more about reflection brainly.com/question/1526267

<h3>Answer details</h3>

Grade: 9

Subject: mathematics

Chapter: reflection

Keywords: reflection

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Suppose the method of tree ring dating gave the following dates A.D. for an archaeological excavation site. Please show your wor

natita [175]

Answer:

a) $\bar X=\frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444$

b) $s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s = 45.580$

c) For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

d) $1260.444-2.306 \frac{45.580}{\sqrt{9}}= 1225.408$

$1260.444+2.306 \frac{45.580}{\sqrt{9}}= 1295.480$

Step-by-step explanation:

For this case we ave the following data:

1245 1245 1321 1191 1295 1330 1239 1250 1228

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got:

$\bar X= \frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444$

Part b

For this case we can use the following formula in order to find the sample standard deviation:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s = 45.580$

Part c

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

Part d

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

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

We need to find the degrees of freedom given by:

$df = n-1 = 9-1=8$

Since the Confidence is 0.95 or 95%, the value of $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=2.306$