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3 years ago

I need help with this

Mathematics

2 answers:

EastWind [94]3 years ago

6 0

I think you add the numbers together, and then divide by three

Art [367]3 years ago

5 0

What is it?

Do you have a better picture.

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Help asap pls for brainliest!

Brut [27]

Answer:

Step-by-step explanation:

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2 years ago

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At a school number of boys : number of girls = 11 : 9

aliya0001 [1]

Ok, so 11:9
11-9 = 2
which means that 124= 2:1
124/2 = 62
so 62:0
now 11 x 62= 682
9 x 62= 558

so the ratio is 682:558
now 682 + 558 = 1,240
There are 1,240 total students in the school. 682 boys and 558 girls

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2 years ago

Emily has $110 in a savings account this year. She plans on depositing $10 into the savings account each following year. How muc

Masteriza [31]

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

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2 years ago

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Where are the Holy Lands located? Why is this area so important to 3 major religions?

nekit [7.7K]

It is located in Wisconsin, I don't know the second question

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3 years ago

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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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2 years ago