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zimovet [89]
3 years ago
9

Write the equation of the line passing through point (-2 - 2) with a slope of -1

Mathematics
1 answer:
sukhopar [10]3 years ago
4 0
I think that y=-1x-4 is the answer
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Look at this triangle.<br> 8 cm<br> 22 cm<br> Work out length AB.
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Complete the square for the expression, then factor the trinomial.
Nuetrik [128]

Answer:

Completing the square we get: (x-2)^2-(2)^2

and factoring the term we get: x(x-4)

Step-by-step explanation:

We need to complete the square for the expression x^2-4x

For completing the square the expression would be of form a^2+2ab+b^2=(a+b)^2

For given expression we have to add (2)^2 and subtract to make it complete the square.

x^2-2(x)(2)+(2)^2-(2)^2\\(x-2)^2-(2)^2

Now, we have to factor the polynomial using formula (a)^2-(b)^2=(a-b)(a+b)

So, (x-2)^2-(2)^2\\=(x-2-2)(x-2+2)\\=(x-4)(x-0)\\=x(x-4)

Completing the square we get: (x-2)^2-(2)^2

and factoring the term we get: x(x-4)

5 0
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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