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Sauron [17]
3 years ago
9

HELP PICS INCLUDED! WILL GIVE BRAINLIEST! SHOW WORK AND EXPLAIN

Mathematics
1 answer:
Naddik [55]3 years ago
4 0

I assume you know about the dot product, and that for two vectors \mathbf a and \mathbf b, the angle between them \theta satisfies

\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta\iff\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{\|\mathbf a\|\|\mathbf b\|}

Then the vectors are parallel if the angle between them is 0 or 180 degrees (0 or pi radians), which would make \cos\theta=1 or \cos\theta=-1, respectively.

Part A)

\vec v_1=\langle\sqrt3,1\rangle\implies\|\vec v_1\|=\sqrt{(\sqrt3)^2+1^2}=\sqrt4=2

\vec v_2=\langle-\sqrt3,-1\rangle=-\vec v_1\implies\|\vec v_2\|=\|\vec v_1\|=2

\vec v_1\cdot\vec v_2=(\sqrt3)(-\sqrt3)+(1)(-1)=-4

Then the angle between \vec v_1,\vec v_2 is such that

\cos\theta=\dfrac{-4}{(2)(2)}=-1\implies\theta=\pi\,\mathrm{rad}

so these vectors are parallel ("antiparallel", more specifically, which means they are parallel but point in opposite directions).

Part B) involves the same computations:

\vec u_1=\langle2,3\rangle\implies\|\vec u_1\|=\sqrt{2^2+3^2}=\sqrt{13}

\vec u_2 has the same components but differing by sign and order, as \vec u_1; its magnitude remains the same, though:

\vec u_2=\langle-3,-2\rangle\implies\|\vec u_2\|=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}

\vec u_1\cdot\vec u_2=(2)(-3)+(3)(-2)=-12

\implies\cos\theta=\dfrac{-12}{(\sqrt{13})(\sqrt{13})}=-\dfrac{12}{13}\implies\theta=\cos^{-1}\left(-\dfrac{12}{13}\right)

which is neither 0 nor pi, which means these vectors are not parallel.

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Answer:

Part 1) <em>Sphere</em> The surface area is equal to SA=196\pi\ m^{2} and the volume is equal to V=\frac{1,372}{3}\pi\ m^{3}

Part 2) <em>Cone</em> The surface area is equal to SA=(16+4\sqrt{65})\pi\ units^{2} and the volume is equal to V=\frac{112}{3}\pi\ units^{3}

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Step-by-step explanation:

Part 1) The figure is a sphere

a) Find the surface area

The surface area of the sphere is equal to

SA=4\pi r^{2}

we have

r=14/2=7\ m ----> the radius is half the diameter

substitute

SA=4\pi (7)^{2}

SA=196\pi\ m^{2}

b) Find the volume

The volume of the sphere is equal to

V=\frac{4}{3}\pi r^{3}

we have

r=14/2=7\ m ----> the radius is half the diameter

substitute

V=\frac{4}{3}\pi (7)^{3}

V=\frac{1,372}{3}\pi\ m^{3}

Part 2) The figure is a cone

a) Find the surface area

The surface area of a cone is equal to

SA=\pi r^{2} +\pi rl

we have

r=4\ units

h=7\ units

Applying Pythagoras Theorem find the value of l (slant height)

l^{2}=r^{2} +h^{2}

substitute the values

l^{2}=4^{2} +7^{2}

l^{2}=65

l=\sqrt{65}\ units

so

SA=\pi (4)^{2} +\pi (4)(\sqrt{65})

SA=16\pi +4\sqrt{65}\pi

SA=(16+4\sqrt{65})\pi\ units^{2}

b) Find the volume

The volume of a cone is equal to

V=\frac{1}{3}\pi r^{2}h

we have

r=4\ units

h=7\ units

substitute

V=\frac{1}{3}\pi (4)^{2}(7)

V=\frac{112}{3}\pi\ units^{3}

Part 3) The figure is a triangular prism

a) The surface area of the triangular prism is equal to

SA=2B+PL

where

B is the area of the triangular base

P is the perimeter of the triangular base

L is the length of the prism

<em>Find the area of the base B</em>

B=\frac{1}{2} (2.7)(2.3)=3.105\ mm^{2}

<em>Find the perimeter of the base P</em>

P=2.7*3=8.1\ mm

we have

L=5.6\ mm

substitute the values

SA=2(3.105)+(8.1)(5.6)=51.57\ mm^{2}

b) Find the volume

The volume of the triangular prism is equal to

V=BL

where

B is the area of the triangular base

L is the length of the prism

we have

B=3.105\ mm^{2}

L=5.6\ mm

substitute

V=(3.105)(5.6)=17.388\ mm^{3}

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