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3 years ago

What is 3/4 times 2/4

Mathematics

1 answer:

Sonja [21]3 years ago

5 0

Answer:

6/16, which can be reduced to 3/8

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Solve for x.

AfilCa [17]

You need to put the numbers it give you, algo will am I supposed to solve that

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3 years ago

A student has earned a 92%, 88%, 65%, 79%, and 99% on her unit tests. What is the student's mean unit test score? Enter your ans

Feliz [49]

92+88+65+79+99=423
423/5=84.6%
5 is from the amount of percentages given

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3 years ago

Read 2 more answers

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

Solve this equation using the radical equation method

Yuki888 [10]

The domain:
$3x+6\geq0\ \ |-6\\3x\geq-6\ \ \ |:3\\x\geq-2\\\\D:x\in\left< -2;\ \infty\right)$

$2x=\sqrt{3x+6}-1\ \ \ |+1\\\\2x+1=\sqrt{3x+6}\ \ \ \ |^2\\\\(2x+1)^2=\left(\sqrt{3x+6}\right)^2\ \ \ |use:\ (a+b)^2=a^2+2ab+b^2\\\\(2x)^2+2\cdot2x\cdot1+1^2=3x+6\\\\4x^2+4x+1=3x+6\ \ \ \ |-3x-6\\\\4x^2+x-5=0\\\\4x^2-4x+5x-5=0\\\\4x(x-1)+5(x-1)=0\\\\(x-1)(4x+5)=0\iff x-1=0\ \vee\ 4x+5=0\\\\x=1\ \vee\ 4x=-5\\\\x=1\in D\ \vee\ x=-1.25\in D$

$Answer:\ \boxed{x=-1.25\ \vee\ x=-1}$

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3 years ago

Please help!!!!!!!!!!!!!!! i need this answer really fast!!!

yan [13]

Answer:

What's the question you f... you amazing human being

3 0

1 year ago