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3 years ago

Given the inequality 2y + 6 > 20 Graph a solution

Mathematics

2 answers:

DedPeter [7]3 years ago

3 0

Answer:

y > 7

Step-by-step explanation:

2y > 20-6

2y > 14

2y/2 > 14/2

y > 7

gayaneshka [121]3 years ago

3 0

Answer: y>y (graphing see below)

Step-by-step explanation:

2y+6>20

2y>14

y>7

Hope this helps!! :)

Please let me know if you have any question

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A mass weighing 16 pounds stretches a spring (8/3) feet. The mass is initially released from rest from a point 2 feet below the

mezya [45]

Answer with Step-by-step explanation:

Let a mass weighing 16 pounds stretches a spring $\frac{8}{3}$ feet.

Mass= $m=\frac{W}{g}$

Mass= $m=\frac{16}{32}$

$g=32 ft/s^2$

Mass,m= $\frac{1}{2}$ Slug

By hook's law

$w=kx$

$16=\frac{8}{3} k$

$k=\frac{16\times 3}{8}=6 lb/ft$

$f(t)=10cos(3t)$

A damping force is numerically equal to 1/2 the instantaneous velocity

$\beta=\frac{1}{2}$

Equation of motion :

$m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)$

Using this equation

$\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)$

$\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)$

$\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)$

Auxillary equation

$m^2+m+12=0$

$m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}$

$m=\frac{-1\pmi\sqrt{47}}{2}$

$m_1=\frac{-1+i\sqrt{47}}{2}$

$m_2=\frac{-1-i\sqrt{47}}{2}$

Complementary function

$e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})$

To find the particular solution using undetermined coefficient method

$x_p(t)=Acos(3t)+Bsin(3t)$

$x'_p(t)=-3Asin(3t)+3Bcos(3t)$

$x''_p(t)=-9Acos(3t)-9sin(3t)$

This solution satisfied the equation therefore, substitute the values in the differential equation

$-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)$

$(3B+3A)cos(3t)+(3B-3A)sin(3t)=20cso(3t)$

Comparing on both sides

$3B+3A=20$

$3B-3A=0$

Adding both equation then, we get

$6B=20$

$B=\frac{20}{6}=\frac{10}{3}$

Substitute the value of B in any equation

$3A+10=20$

$3A=20-10=10$

$A=\frac{10}{3}$

Particular solution, $x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)$

Now, the general solution

$x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)$

From initial condition

x(0)=2 ft

x'(0)=0

Substitute the values t=0 and x(0)=2

$2=c_1+\frac{10}{3}$

$2-\frac{10}{3}=c_1$

$c_1=\frac{-4}{3}$

$x'(t)=-\frac{1}{2}e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+e^{-\frac{t}{2}}(-c_1\frac{\sqrt{47}}{2}sin(\frac{\sqrt{47}t}{2})+\frac{\sqrt{47}}{2}c_2cos(\frac{\sqrt{47}t}{2})-10sin(3t)+10cos(3t)$

Substitute x'(0)=0

$0=-\frac{1}{2}\times c_1+10+\frac{\sqrt{47}}{2}c_2$

$\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0$

$\frac{\sqrt{47}}{2}c_2=-\frac{2}{3}-10=-\frac{32}{3}$

$c_2==-\frac{64}{3\sqrt{47}}$

Substitute the values then we get

$x(t)=e^{-\frac{t}{2}}(-\frac{4}{3}cos(\frac{\sqrt{47}t}{2})-\frac{64}{3\sqrt{47}}sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)$

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snow_tiger [21]

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Step-by-step explanation:

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Hitman42 [59]

<h3>Answer:</h3>

x = 3

<h3>Explanation:</h3>

The product of the lengths of segments from the intersection point to the circle is the same for both secants.

... 1×6 = 2×x

... 6/2 = x = 3 . . . . . divide by 2

_____

<em>Comment on secant geometry</em>

Interestingly, this relation is true whether the point of intersection of the secants is inside the circle or outside.

When it is outside, the product is of the distance to the near intersection with the circle and the distance to the far intersection with the circle.

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Step-by-step explanation:

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