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Neko [114]
3 years ago
11

PLEASE HELP IT’S DUE TMR

Mathematics
1 answer:
irakobra [83]3 years ago
4 0
The anwser to the first one is 0.6
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I need help with this problem it’s due at midnight and I’m so tired of trying. Thanks it’s number 4 btw
bearhunter [10]

Answer:

Look below

Step-by-step explanation:

A.

8 + 9 + 10 = 27

9 + 10 + 11 = 30

10 + 11 + 12 = 33

B. Every answer is divisible by 3, every consecutive answer is also 3 more than the one before it.

C. Yes

19 + 20 + 21 = 60

D. No

8 is not divisible by 3. Also, the closest you can get is 6 (1 + 2 + 3) or 9 (2 + 3 + 4).

I hope this helped! If you need anything else, let me know in the comments. :)

6 0
3 years ago
Can someone help? Please explain how you got your answer.
frez [133]
The tablecloth has 42" diameter.
Add 8" at each end of the diameter to get 42" + 8" + 8" = 58".
The tablecloth has 58" diameter.
The radius is half the diameter.
Radius = diameter/2 = 58"/2 = 29".
Answer: The radius of the tablecloth is 29".

3 0
3 years ago
Read 2 more answers
Keith thinks of a number. When he multiples the number by 6 and subtracts 19.85 from the product, he gets 29.77. Find the number
kirill115 [55]

let the number be X

6x-19.85=29.77

6x=29.77+19.85

6x=49.62

x=8.27

Hope that helps :)

-Asmaa Ghazzawi

8 0
3 years ago
What team has not been to a super bowl
mylen [45]
All of the above easy one
5 0
3 years ago
Read 2 more answers
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
3 years ago
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