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mafiozo [28]
2 years ago
13

Prove that:(cosecA-sinA)(secA-cosA)(tanA+cotA)=1

Mathematics
1 answer:
dedylja [7]2 years ago
4 0

Answer:

.............................

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Anthony has to practice his guitar for 1.75 hours each morning and 2.5 hours each evening for the first 4 days this week. How ma
weqwewe [10]

Answer:

17 hours

Step-by-step explanation:

From the above question,

Anthony is practicing 4 days in a week

Hence

The number of hours he practices in the morning for 4 days =

1.75 hours × 4

= 7 hours

The number of hours he practices in the evening for 4 days =

2.5 hours × 4

= 10 hours

Therefore, the total number of hours he practices this week

= 7 hours + 10 hours

= 17 hours

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2 years ago
What is 300 as a decimal
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Answer:

300.00

Step-by-step explanation:

6 0
3 years ago
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Estimate 1.133+7.684+4.718
avanturin [10]

Answer:

1.133 is about 1, 7.684 is about 8, and 4.718 is about 5. Thus, 1 + 8 + 5 = about 14.

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2 years ago
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Question 2; how do i solve it?
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Please excuse my dear aunt sally or pemdas
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3 years ago
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.

6 0
2 years ago
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