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2 years ago

Please find it please

Mathematics

2 answers:

rusak2 [61]2 years ago

8 0

The answer is 0.225 ur welcome (u can use a fraction calculator online)

Sliva [168]2 years ago

6 0

Answer:

Step-by-step explanation:

cross multiplication

8/5 = 1.6

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A rectangle with width of 4. There is a vertical line that splits rectangle into two sections with different lengths. One length

Serhud [2]

Answer:

<h2>4x+8</h2><h2>Step-by-step explanation:</h2>

If you distribute 4(x+2) using the distributive property, which states that multiplying each number individually by a same # is the same as multiplying the #s' sum by that #.

Thus, 4*(x+2) is the same as 4*x+4*2 which simplifies to 4x+8, Option 2

6 0

4 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

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

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3 years ago

-1/6b=3<br><br> b=<br><br> A. -18<br> B. -2<br> C 2<br> D 18

goldenfox [79]

$-\frac{1}{6}b = 3\\\\-\frac{1}{6}b\times -6 = 3\times -6\\\\ b = -18$

7 0

4 years ago

I need to know what to do and the answer

Anni [7]

You would take 180•15% to get 27 then you would subtract 180-27 to get 153. After you do the first you would do 200•20% to get 40 200-40=160 so the cheaper one would be the 180 because about 15% discount it’s 153 and 160>153

5 0

3 years ago

Make r the subject of the formula in I=E/R+r

allochka39001 [22]

Step-by-step explanation:

$I = \frac{E }{ R} + r$