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2 years ago

Please answer fast!

Mathematics

1 answer:

pentagon [3]2 years ago

5 0

Answer:

4 and 2

Step-by-step explanation:

a positive times a negative makes a negative

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DIA [1.3K]

The correct answer is D, -5 1/3.

The reason is that on a number line -3 1/4 would be placed to the right of -5 1/3.

The further you are to the right on the number line, the bigger the number.

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2 years ago

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Mrrafil [7]

Answer:

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Step-by-step explanation:

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2 years ago

The lunch lady has 3 pounds of lasagna left over. If she makes 1/6-pound servings, how many servings of

maksim [4K]

Answer:

⇒ $\frac{1}{6}$ ×3

⇒ $\frac{1}{2}$

Step-by-step explanation:

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2 years ago

HELPPP ASAP!!!<br><br> How many cubes are in this set?

Stella [2.4K]

Answer:

Step-by-step explanation:

7 because it's not asking for squares it asking for cubes

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2 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