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3 years ago

What is 66/100 in simplest form

Mathematics

2 answers:

nikdorinn [45]3 years ago

6 0

The answer is 33/50 because you can divide both 100 and 66 by 2

damaskus [11]3 years ago

5 0

The answer to your question is 33/50 cause that is what I found on goggle

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Which one of these relationships is different than the other three? <br> A <br> B <br> C<br> D

wlad13 [49]

Answer:

B differs from the others

Step-by-step explanation:

The other lines all have slopes of 5. Graph b has a slope of 5.5

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2 years ago

Math help pls all 4 questions please

zvonat [6]

A. 32500
b. 60,400
c. 2.4 x 10 ^ -6
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3 years ago

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Plz read the question then ANWSER then tell me what u did, to get the amwser/ how u did it

pshichka [43]

Answer:

Step-by-step explanation: use Triangle formula Area= 1/2xbxh

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3 years ago

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Find the equation of the line that is perpendicular to y=3/4x-3 and passes through the point (3,-2)

morpeh [17]

Answer:

Step-by-step explanation:

perp. -4/3

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y + 2 = -4/3x + 4

y = -4/3x + 2

4 0

2 years ago

Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel

nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

$\theta=\frac{\pi}{3}$

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be= $r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j$

By using $cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Force along the plane will be= $\mid F_x\mid=F\cdot r$

Force along the plane will be = $\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3$ N

By using $i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0$

Therefore, force along the plane= $\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)$

2.The vector perpendicular to the plane= $r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j$

The force perpendicular to the plane= $\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

The force perpendicular to the plane= $4$ N

Therefore, $F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $2\sqrt 3i-6j-2\sqrt3 i-2j=-8j$

Hence,sum of two component of forces=Total force.

6 0

3 years ago