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daser333 [38]
3 years ago
6

What is 10*10/11 simplified to a mixed number

Mathematics
2 answers:
g100num [7]3 years ago
8 0

Answer:

9 1/11

Step-by-step explanation:

10*10= 100

100/11

9 and 1/11

Sunny_sXe [5.5K]3 years ago
8 0

Answer:

it is already a mixed number

Step-by-step explanation:

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∠DBC = 19° since the two angles have to add up to 180, if x = 23, then 23 -4 = 19 for the right, and 7 x 23 = 161, and 161 + 19 = 180.
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Helppp plzzz idk I need help​
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Step-by-step explanation:

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3 years ago
EASY POINTS EVALUATION
dedylja [7]

Answer:

\frac{4}{9}

Step-by-step explanation:

( \frac{16 {}^{ \frac{1}{2} } }{81 {  }^{ \frac{1}{2} } }) \\ (  \frac{ {4}^{2 \times  \frac{1}{2} } }{9^{2 \times  \frac{1}{2} } }) \\ ( \frac{4^{1} }{9 ^{1} } ) \\  \frac{4}{9}

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Read 2 more answers
A spring is oscillating so that its length is a sinusoidal function of time. Its length varies from a minimum of 10 cm to a maxi
Hoochie [10]
Let the x(t) represent the motion of the spring as a function of time, t.

The length of the oscillating spring varies from a minimum of 10 cm to a maximum of 14 cm.
Therefore its amplitude is A = (14 - 10)/2 = 2.

When t = 0 s, x = 12 cm.
Therefore the function is of the form
x(t) = 2 sin(bt) + 12

At t=0, x(t) is decreasing, and it reaches its minimum value when t = 1.2 s.
Therefore, a quarter of the period is 1.2 s.
The period is given by
T/4 = 1.2
T = 4.8 s

That is,
b = (2π)/T = (2π)/4.8 = π/2.4 = 1.309

The function is
x(t) = 2 sin(1.309t) + 12
A plot of x(t) is shown below.

When x(t) = 13.5, obtain
2 sin(1.309t) + 12 = 13.5
sin(1.309t) = (13.5 - 12)/2 = 0.75
1.309t = sin⁻¹ 0.75 = 0.8481 or π - 0.8481
t = 0.8481/1.309 or t = (π - 0.8481)/1.309
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The difference in t is 1.751 - 0.649 = 1.1026.

This difference occurs twice between t=0 and t=8 s.
Therefore the spring length is greater than 13.5 cm for 2*1.1026 = 2.205 s.

Answer:
Between t=0 and t=8, the spring is longer than 13.5 cm for 2.205 s.

8 0
3 years ago
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Taya2010 [7]

Yes I believe they are 1 to 15. If I'm not wrong please dont report me.

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