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2 years ago

Please answer I will make you Brainliest if it's right How do I do this step by step??

Mathematics

2 answers:

lyudmila [28]2 years ago

5 0

Answer:

Step-by-step explanation:

Sav [38]2 years ago

3 0

Answer:

$\sqrt{2 \times 2 \times 2 \times 3 } = \sqrt{24} = \sqrt{4 \times 6} = 2 \sqrt{6}$

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

Gelneren [198K]

Answer:

Step-by-step explanation:

Initial figure times scale factor = result

10 times 2 = 20

I hope this helps!

8 0

1 year ago

If Vector u=(5,-7) and v=(-11,3), 2v-6u=_____ and ||2v-6u||≈_____

Sauron [17]

<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:

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

5 0

2 years ago

Read 2 more answers

Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

lions [1.4K]

Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}$

$(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill$

$\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}$