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3 years ago

A circle has a radius of 6 in. Find the area of its inscribed equilateral triangles.

Mathematics

1 answer:

Lina20 [59]3 years ago

8 0

The area of the inscribed triangle = 3(6^2 x sin 120°) = 3(36sin 120°) = 3(31.18) = 93.53 square in.

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Avoiding an accident while driving can depend on reaction time. Suppose that reaction time, measured from the time the driver fi

Nadya [2.5K]

Answer:

2.5% of drivers have a reaction time more than 2.14 seconds

16% of drivers have a reaction time less than 1.78 seconds

84% of drivers have a reaction time less than 2.02 seconds

Step-by-step explanation:

The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

The normal distribution is symmetric, which means that 50% of the measures are below the mean and 50% are above.

In this problem, we have that:

Mean = 1.9s

Standard deviation = 0.12

What percentage of drivers have a reaction time more than 2.14 seconds?

2.14 = 1.9 + 2*0.12

So 2.14 is two standard deviations above the mean.

Of the 50% of the measures above the mean, 95% are within 2 standard deviations of the mean, so, below 2.14. The other 5% is above.

0.05*0.5 = 0.025

2.5% of drivers have a reaction time more than 2.14 seconds

What percentage of drivers have a reaction time less than 1.78 seconds?

1.78 = 1.9 - 0.12

So 1.78 is one standard deviation below the mean.

Of the 50% of the measures that are below the mean, 68% are within one standard deviation of the mean, that is, greater than 1.78.

100 - 68 = 32

0.32*50 = 0.16

16% of drivers have a reaction time less than 1.78 seconds

What percentage of drivers have a reaction time less than 2.02 seconds?

2.02 = 1.9 + 0.12

So 2.02 is one standard deviation above the mean.

Of the measures that are below the mean, all are below 2.02.

Of those that are above, 68% are below 2.02.

0.5 + 0.68*0.5 = 0.84

84% of drivers have a reaction time less than 2.02 seconds

5 0

3 years ago

Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2

kifflom [539]

Answer:

The solution of the differential equation is $y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

<u>i) Using characteristic equation:</u>

The characteristic equation method assumes that y(t)= $e^{rt}$ , where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

$y'=re^{rt}$

$y"=r^{2}e^{rt}$

$r^{2}e^{rt}+e^{rt}=0$

$(r^{2}+1)e^{rt}=0$

As $e^{rt}$ could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

$y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}$

Using Euler's formula:

$y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]$

$y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)$

$y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)$

The particular solution of the differential equation is given by:

$y(t)_{p}=ASin(2t)+BCos(2t)$

$y'(t)_{p}=2ACos(2t)-2BSin(2t)$

$y''(t)_{p}=-4ASin(2t)-4BCos(2t)$

So we use these derivatives in the differential equation:

$-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)$

$-3ASin(2t)-3BCos(2t)=Sin(2t)$

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

$y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)$

Using the initial conditions we can check that C1=5/3 and C2=2

<u>ii) Using Laplace Transform:</u>

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

ℒ[Sin(2t)]= $\frac{2}{(s^{2}+4)}$

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) = $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)-2s-1= $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)= $\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}$

Y(s)= $\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}$

Y(s)= $\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}$

Using partial franction method:

$\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}$

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

B+D=1

A+4C=8

B+4D=6

The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

So,

Y(s)= $\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}$

Y(s)= $-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}$

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[ $-\frac{1}{3} \frac{2}{s^{2}+4}$ ]-ℒ⁻¹[ $2\frac{s }{s^{2}+1}$ ]+ℒ⁻¹[ $\frac{5}{3}\frac{1}{s^{2}+1}$ ]

$y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

3 0

3 years ago

Systems with three variables. please help

kirza4 [7]

X= 6
Y= 0
Z= -2

CHECK
3x + y -2z =22
3(6) + 0 -2(-2) =22
18 + 0 +4= 22
22=22

x+5y+z =4
6 +5(0)+ -2 =4
6 + 0 +-2 = 4
4=4

x+3z=0
6 + 3(-2) =0
6 -6 = 0
0=0

3 0

3 years ago

Find x [Angles and Segment]

Alex_Xolod [135]

Answer:

8.3 cm

Step-by-step explanation:

The product of lengths to the near and far point of intersection with the circle is the same in all cases:

(7 cm)(7 cm) = (y)(11 cm +y) = (4 cm)(4 cm +x)

Since we're only interested in x, we can divide by 4 and subtract 4:

49 cm² = (4 cm)(4 cm +x)

(49/4) cm = 4 cm +x . . . . . . divide by 4 cm

8.25 cm = x . . . . . . . . . . . . . subtract 4 cm

To the nearest tenth, x = 8.3 cm.

_____

For a tangent segment, the two points of intersection with the circle are the same point, so the product of lengths is the square of the length.

___

The angles depend on the size of the circle, which is not given.

6 0

3 years ago

4. Solve the system using substitution.

Yuri [45]

Answer:

Step-by-step explanation:

-3+2=-1 and (9)- 8=1

3 0

3 years ago