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4 years ago

Find the distance between the points (0,0) and(6,-8)

Mathematics

1 answer:

Y_Kistochka [10]4 years ago

7 0

The formula of a distance between two points:

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

We have the points (0, 0) and (6, -8). Substitute:

$d=\sqrt{(6-0)^2+(-8-0)^2}=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=\boxed{10}$

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Which expression is equivalent to 2/9 x 3/4

weeeeeb [17]

Answer:

0.16666666666667

Step-by-step explanation:

Let me know if this is correct.

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3 years ago

Read 2 more answers

Bishwant and sunayana cimplete a oiece of a work in 6 days working together.if bishwant alime can complete it n 10 days.in how m

alina1380 [7]

Step-by-step explanation:

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3 years ago

What is the final cost of a $40 item after a 8% tax is applied

zheka24 [161]

$43.2. to do this, you first find 8% of 40 by doing 40 times 0.08. this is equal to 3.2. next, since it's 8% tax, you add 3.2 to 40. this is how i got $43.2

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3 years ago

How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 10% antifreeze?

musickatia [10]

Answer:

Let x = amount of water to be added with 0% antifreeze

1 = amount of pure antifreeze at 100%

x+1 = amount of the mixture at 60% antifreeze

Step-by-step explanation:

1= .60x + .60

Subtract .60 from each side:

1-.60 = .60x

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Divide both sides by .60

Check: Total antifreeze = 1 gallon in total mixture of 1 + 2/3 gallon = 5/3 gallons. 1 gallon divided by 5/3 equals 3/5 which is 60%. It checks.

5 0

3 years ago

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 $y = e^x$

-----------------------

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.

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3 years ago