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3 years ago

The endpoints of a diameter of a circle are A(3,1) and B(8,13). Find the area of the circle in terms of pi.

Mathematics

1 answer:

Talja [164]3 years ago

7 0

Answer:

A = 169π/4

Step-by-step explanation:

From these two given points we find the length of the diameter of this circle, using the Pythagorean Theorem:

d^2 = (8 - 3)^2 + (13 - 1)^2, or

d^2 = 25 + 144 = 169

Thus, the diameter, d, of this circle is √169, or 13.

The radius is thus 13/2 units.

The area formula is A = πr^2. Here the area is

A = π(169/4), or A = 169π/4

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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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3 years ago

Which is bigger, 7/8" or 9/16"?

Iteru [2.4K]

7/8 and 9/16

7*2/8*2 = 14/16

14/16 and 9/16

Now since we have a common denominator, we can compare the numerators.

14/16 is greater, so 7/8 is greater than 9/16

Answer: 7/8 is bigger

5 0

3 years ago

Read 2 more answers

Which of the following inequality statements best represent the verbal expression?

Alex73 [517]

3rd option.

at least shows that the number can be equal to 2, and also that 2 should be the smallest possible value of the number.

3 0

3 years ago

Need brief explanation about why false is correct

anyanavicka [17]

We are given the equation:

(a + b)! = a! + b!

Testing the given equation

In order to test it, we will let: a = 2 and b = 3

So, we can rewrite the equation as:

(2+3)! = 2! + 3!

5! = 2! + 3!

We know that (5! = 120) , (2! = 2) and (3! = 6):

120 = 2 + 6

We can see that LHS ≠ RHS,

So, we can say that the given equation is incorrect

6 0

2 years ago

What is the answer to 11/12×6=

Mademuasel [1]

Answer:

$\frac{11}{2}$

Step-by-step explanation:

$\frac{11}{12}$ x 6 = $\frac{11}{2}$

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3 years ago

The endpoints of a diameter of a circle are ​A(3​,​1) and ​B(8​,13​). Find the area of the circle in terms of pi.

The endpoints of a diameter of a circle are A(3,1) and B(8,13). Find the area of the circle in terms of pi.