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3 years ago

Show that the points A (-3, 2), B (-6, 4) and C (1, 8) are vertices of a right triangle.

Mathematics

1 answer:

galben [10]3 years ago

5 0

Answer:

See below.

Step-by-step explanation:

For the triangle to be a right triangle there must be a pair adjacent sides which are at right angles to each other - that is whose slope product = -1.

Slope of AB = (4-2)/(-6- -3) = -2/3.

Slope of BC = (8-4)/ (1 - - 6) = 2/7

Slope of AC = (8-2) / (1 - -3) = 6/4 = 3/2.

Now 3/2 * -2/3 = -1 so sides AB and AC are at right angles and the 3 points are the vertices of a right triangle.

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K.Brew sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. R

Elan Coil [88]

Answer:

Step-by-step explanation:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = mean sale amount for mail order sales = 82.70

x2 = mean sale amount for internet sales = 66.9

s1 = sample standard deviation for mail order sales = 16.25

s2 = sample standard deviation for internet sales = 20.25

n1 = number of mail order sales = 17

n2 = number of internet sales = 10

For a 99% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (17 - 1) + (10 - 1) = 25

z = 2.787

Margin of error =

z√(s²/n1 + s2²/n2) = 2.787√(16.25²/17 + 20.25²/10) = 20.956190

4 0

3 years ago

Select the diagram showing a pair of alternate interior angles:

Nat2105 [25]

The orange one is an example of alternate interior angles.

4 0

2 years ago

6.24 x 10-5 - 4.091 x 103 =

castortr0y [4]

Is the 10 taking away 5 or was it meant to me 10.5 ?

6 0

3 years ago

Read 2 more answers

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

Rename as a mixed number in simplest form. 15 22/20

Semenov [28]

16 1/10
Because you have 15 and then for 22/20 it is a improper fraction so take 20 away from it to get 2/10 then add 1 to 15 for 16 and then simplify 2/20 to 1/10

8 0

3 years ago

Read 2 more answers