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svlad2 [7]
3 years ago
7

A teacher asked her pupils what their favorite school subject was 3 pupils said PE was their favorite subject How many said that

maths was their favorite subject
HELP PLZZZZZZZZ
SOMEONE
Mathematics
2 answers:
pychu [463]3 years ago
8 0
What are the number of students in the class then only I give answer
alisha [4.7K]3 years ago
6 0

Answer:

None of the students said maths was his/her favourite subject

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Around 30 don’t work
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3/4×-18=1/4×-4 help me slove
WINSTONCH [101]

3/4 x -18 = 1/4 x -4

subtract 1/4 x from each side

3/4 x -1/4 x -18 = 1/4 x- 1/4 x -4

2/4 x - 18 = -4

add 18 to each side

1/2 x -18 + 18 = -4 + 18

1/2 x =14

multiply by 2 on each side

2* 1/2 x = 14 * 2

x =28

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A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th
Yuri [45]

Answer:

15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

y(0)=0, y'(0)=0

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: F_{r} (t) that correspond to the force of resistance on the mass by the action of the spring and F(t) that is an external force with unknown direction (that does not specify in the enounce).

For determinate F_{r} (t) we can use Hooke's Law given by the formula F_{r} (t) = k y(t) where k correspond to the elastic constant of the spring and y(t) correspond to  the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m}  = 441.45 \frac{N}{m}

Now we apply Newton's 2nd Law and obtaint that...

F_{r} (t) ± F(t) = ma(t)

F_{r} (t) = ky(t) = 441.45y(t)

F(t) = 170 cos(5t)

m = 15 kg

a(t) = \frac{d^{2}y(t)}{dt^{2} }

Finally... 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

We know from the problem that there's not initial displacement and initial velocity, so... y(0)=0 and y'(0)=0

Finally the Initial Value Problem that models the situation describe by the problem is

\left \{ 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.

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What's the sum of the number m and the number after it?
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How is anyone supposed to answer this?
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