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3 years ago

Find the length of the segment with endpoints of (3, 2) and (-3, -6).

Mathematics

1 answer:

nevsk [136]3 years ago

8 0

The distance between (3, 2) and (-3, -6) is 10 units

The correct answer is B) 10

Further explanation:

Given points are:

(x1,y1) = (3,2)

(x2,y2)=(-3,-6)

The distance formula is given by:

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

Putting the values

$d=\sqrt{(-3-3)^2+(-6-2)^2}\\=\sqrt{(-6)^2+(-8)^2} \\=\sqrt{36+64}\\ =\sqrt{100}\\d=10$

The distance between (3, 2) and (-3, -6) is 10 units

Keywords: Magnitude, Distance

Learn more about magnitude at:

#LearnwithBrainly

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Answer:

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Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

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$dx=(t-sint)dt$

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where k is a constant due to integration. With x(0)=1, substitute:

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Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

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Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

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Finally:

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serious [3.7K]

Answer:

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Step-by-step explanation:

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