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Ipatiy [6.2K]
3 years ago
7

I need help on number 19.

Mathematics
2 answers:
scoundrel [369]3 years ago
8 0
C it would be C becuse 3 8ths plus 1 4ths equals 4 8ths
8_murik_8 [283]3 years ago
6 0
Add the fractions 1/4 and 3/8
First find the LCD of 1/4 and 3/8.  8 is the LCD.
Turn 1/4 into 2/8.
Now add.
3/8+2/8=5/8

Hope this helps!
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Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En
ArbitrLikvidat [17]

Answer:

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, c) x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right), y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right).

Step-by-step explanation:

The equation of the circle is:

x^{2} + (y-1)^{2} = 16

After some algebraic and trigonometric handling:

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t

Where:

\frac{x}{4} = \cos t

\frac{y-1}{4} = \sin t

Finally,

x = 4\cdot \cos t

y = 1 + 4\cdot \sin t

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

c) x = 4\cdot \cos t'', y = 1 + 4\cdot \sin t''

Where:

4\cdot \cos t' = 0

1 + 4\cdot \sin t' = 5

The solution is t' = \frac{\pi}{2}

The parametric equations are:

x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right)

y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)

7 0
3 years ago
Austin has 125% more books than his younger sister. If his sister has 24 books, how many books does he have? Write your proporti
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Answer:

30 books

Step-by-step explanation:

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Step-by-step explanation:

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Please help. Ella has 5 pink bows, 1 blue bow, and 4 purple bows in a box, she will randomly choose 1 bow from the box. What is
pogonyaev

Answer:

4/10 chance. Or 2/5. 40% chance.

Step-by-step explanation:

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I have 3 digits. If I round to the nearest hundred, I am 200. Multiply my ones digit by 2 and you get my tens digit. Both my one
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Answer:

the answer is 393

Step-by-step explanation:

hope we can be friends

can i please get brainliest

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