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3 years ago

Find the distance between the two points (0,0), (4,3)

Mathematics

1 answer:

Amiraneli [1.4K]3 years ago

4 0

Answer:

Distance between the points=5

Step-by-step explanation:

The distance between two points in coordinate geometry is measured by using the distance formula:

If we have given two points

$(x_1,y_1)\\\\(x_2,y_2)$

Distance formula= $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

For (0,0) and (4,3)

Distance:

$\sqrt{(4-0)^2+(3-0)^2} \\\\\sqrt{4^2+3^2}\\\\\sqrt{16+9}\\\\\sqrt{25}$

Distance between the points=5

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Given the slope a slope of 2 and a y-intercept of 7, write the equation in slope intercept form

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Step-by-step explanation:

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2 years ago

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2 years ago

Read 2 more answers

Find sin D sin E cos D and cos E

Andrew [12]

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Answer:

sin(D) = cos(E) = (√3)/2

cos(D) = sin(E) = 1/2

Step-by-step explanation:

The mnemonic SOH CAH TOA is intended to remind you of the relationships between trig functions and right triangle sides.

Sin = Opposite/Hypotenuse

Cos = Adjacent/Hypotenuse

For this diagram, this means ...

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2 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago