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3 years ago

Find the distance between two points ( -1,1) and (2,-4)

Mathematics

1 answer:

sladkih [1.3K]3 years ago

3 0

Answer:

The distance between two points ( -1,1) and (2,-4) is:

$d=\sqrt{34}$ or d = 5.8 units.

Step-by-step explanation:

Given the points

(-1, 1)

(2, -4)

Finding the distance between (-1, 1) and (2, -4) using the formula

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

substitute (x₁, y₁) = (-1, 1) and (x₂, y₂) = (2, -4)

$=\sqrt{\left(2-\left(-1\right)\right)^2+\left(-4-1\right)^2}$

$=\sqrt{\left(2+1\right)^2+\left(-4-1\right)^2}$

$=\sqrt{3^2+5^2}$

$=\sqrt{9+25}$

$=\sqrt{34}$ units

$d = 5.8$ units

Therefore, the distance between two points ( -1,1) and (2,-4) is:

$d=\sqrt{34}$ or d = 5.8 units.

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1 year ago

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First find the critical points of f :

$f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7$

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so the point (1, 0) is the only critical point, at which we have

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Next check for critical points along the boundary, which can be found by converting to polar coordinates:

$f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t$

Find the critical points of g :

$\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0$

$\implies\sin t=0\text{ OR }1+5\cos t=0$

$\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi$

where n is any integer. We get 4 critical points in the interval [0, 2π) at

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So f has a minimum of -7 and a maximum of 299.

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Answer:

Step-by-step explanation:

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Variance, σ2: 2.25

Steps

σ2 =
Σ(xi - μ)2
N
=
(0 - 1.5)2 + ... + (5 - 1.5)2
10
=
22.5
10
= 2.25
σ = √2.25
= 1.5

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