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denis23 [38]
2 years ago
11

If the initial figure is 10 and then it is dilated with a scale factor of 2 what's the perimeter of dilated figure?

Mathematics
1 answer:
Gelneren [198K]2 years ago
8 0

Answer:

20

Step-by-step explanation:

Initial figure times scale factor = result

10 times 2 = 20

I hope this helps!

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If Vector u=(5,-7) and v=(-11,3), 2v-6u=_____ and ||2v-6u||≈_____
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<h2>Answer:</h2>

2\vec{v}-6\vec{u}=(-52,48) \\ \\ ||2\vec{v}-6\vec{u}||=75.28}

<h2>Step-by-step explanation:</h2>

In this problem we have two vectors:

\vec{u}=(5,-7) \ and \ \vec{v}=(-11,3)

So we need to find two things:

2\vec{v}-6\vec{u}

and:

||2\vec{v}-6\vec{u}||

FIRST:

In this case we have the multiplication of vectors by scalars. A scalar is a simple number, so:

2\vec{v}-6\vec{u} \\ \\ Replace \ \vec{v} \ and \ \vec{u} \ by \ the \ given \ vectors: \\ \\ 2(-11,3)-6(5,-7) \\ \\ Multiply \ each \ component \ by \ the \ corresponding \ scalar:\\ \\ (2\times (-11),2\times 3)+(-6\times 5,-6\times (-7)) \\ \\ (-22,6)+(-30,42) \\ \\ Sum \ of \ vectors: \\ \\ (-22-30,6+42) \\ \\ \therefore \boxed{(-52,48)}

SECOND:

If we name:

\vec{w}=2\vec{v}-6\vec{u}

Then, ||2\vec{v}-6\vec{u}|| is the magnitude of the vector \vec{w}. Therefore:

||\vec{w}||=||2\vec{v}-6\vec{u}|| \\ \\ ||\vec{w}||=||(-52,48)|| \\ \\ ||\vec{w}||=\sqrt{(-58)^2+48^2} \\ \\ ||\vec{w}||=\sqrt{3364+2304} \\ \\ ||\vec{w}||=\sqrt{5668} \\ \\ \boxed{||\vec{w}||=75.28}

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