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Murrr4er [49]
3 years ago
7

Sam has the lead role in the school play. He still has to memorize 1/2 of his lines. Suppose Sam memorizes 1/3 of his lines toda

y. What fractions of his lines will he have left to memorize?
Mathematics
2 answers:
DaniilM [7]3 years ago
4 0

Answer: 1/6

Step-by-step explanation:

Since Sam has to memorize 1/2 of his lines and he memorizes 1/3 of his lines today, the fractions of his lines that he has left to memorize will be gotten by subtracting 1/3 from 1/2. This will be:

= 1/2 - 1/3

= 3/6 - 2/6

= 1/6

Therefore, he has 1/6 of his lines left to memorize.

poizon [28]3 years ago
4 0

Answer:

1/6

Step-by-step explanation:

The fractions of his lines will he have left to memorize is calculated as:

Number of lines he has to memorize - Number of line he memorized today

Number of lines he has to memorize = 1/2

Number of line he memorized today = 1/3

Hence,

= 1/2 - 1/3

Lowest Common Denominator = 6

= 3 - 2/6

= 1/6

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5 0
4 years ago
Find the radius r of C.
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10

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16 + r = √(24^2 + r^2)

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5 0
3 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 3N

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=4N

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
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