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11 days ago

In △ABC , m∠A=53°,m∠B=17°, and a=27. Find the perimeter of the triangle.

1 answer:

6 0

Utilizing the Law of Sines (sinA/a=sinB/b=sinC/c) and recognizing that the angles in a triangle add up to 180°.

The angle C calculates to 180-53-17=110°

Thus, we have 27/sin53=b/sin17=c/sin110

This leads to b=27sin17/sin53, c=27sin110/sin53

The perimeter is defined as a+b+c, so

p=27+27sin17/sin53+27sin110/sin53 units

p≈68.65 units (rounded to the nearest hundredth of a unit)

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A job shop consists of three machines and two repairmen. The amount of time a machine works before breaking down is exponentiall

tester [8842]

Answer:

Step-by-step explanation:

Define X(t) as the count of machines that are out of order at time t.

The specified issue is modeled with a birth-death process within the confined set

S={0, 1, 2, 3} alongside

$\lambda_0=\frac{3}{10}, \mu_1=\frac{1}{8}\\\\ \lambda_1=\frac{2}{10}, \mu_2=\frac{2}{8}\\\\ \lambda_2=\frac{1}{10}, \mu_3=\frac{2}{8}$

The balancing equations for the birth-death process are given by $\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2$

because the exiting rate = the entering rate

0 $\lambda_0P_0=\mu_1P_1$

1 $(\lambda_1+\mu_1)P_1= \mu_2P_2 + \lambda_0P_0$

2 $(\lambda_2+\mu_2)P_2= \mu_3P_3 + \lambda_1P_1$

$P_1=\frac{12}{5}P_0=P_0=\frac{5}{12}P_1\\\\P_2=\frac{48}{25}P_0=P_0=\frac{25}{48}P_2\\\\P_3=\frac{192}{250}P_0=P_0=\frac{250}{192}P_3$

Since $\sum\limits^3_{i=0} {P_i=1}\\\\p_0=[1+\frac{5}{12}+\frac{48}{25}+\frac{192}{250}]^{-1}=\frac{250}{1522}$

The average number of machines not in operation is reflected in the mean of the stationary distribution $P_1+2P_2+3P_3=\frac{2136}{751}$

The fraction of time both repair technicians are engaged $P_2+P_3=\frac{672}{1522}=\frac{336}{761}$

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23 days ago

An object is launched into the air. The projectile motion of the object can be modeled using the function h(t) = –16t2 + 72t + 5

PIT_PIT [9117]

I'm uncertain if this will be of assistance.

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21 day ago

The annual net income of a company for the period 2007–2011 could be approximated by P(t) = 1.6t2 − 11t + 44 billion dollars (2

babunello [8412]

Answer:

$P'(t) = 3.2 t -11$

To identify the critical point, we set the derivative equal to zero, resulting in:

$3.2 t-11= 0$

$t = \frac{11}{3.2}= 3.4375$

Calculating the second derivative yields:

$P''(t) = 3.2 >0$

This indicates that t = 3.4375 marks the minimum value for the function. Substituting this back into the original function gives us:

$P(3.4375) = 1.6(3.4375)^2 - (11*3.4375) +44 = 25.094$

Thus, the minimum annual income is found at t = 3.43 (between the years 2008 and 2009), with a value of 25.094

Step-by-step explanation:

In this scenario, we have the function:

$P(t) = 1.6 t^2 -11t +44$

Where P represents the annual net income from 2007-2011, and $2 \leq t \leq 7$

t is the number of years since early 2005

To discover the lowest income, we utilize the derivative given by:

$P'(t) = 3.2 t -11$

Setting this derivative to zero allows us to find the critical point, leading to:

$3.2 t-11= 0$

$t = \frac{11}{3.2}= 3.4375$

Calculating the second derivative reveals:

$P''(t) = 3.2 >0$

Therefore, we conclude that t = 3.4375 is the minimum value, substituting into the original function results in:

$P(3.4375) = 1.6(3.4375)^2 - (11*3.4375) +44 = 25.094$

Thus, the minimum annual income occurs at t = 3.43 (between 2008 and 2009) with the value being 25.094

5 0

5 days ago

Wyatt is going to a carnival that has games and rides. Each game costs $1.25 and each ride costs $2.75. Wyatt spent $20.25 altog

Inessa [9000]

Answer:

Wyatt participated in 3 games and enjoyed 6 rides

Step-by-step explanation:

Let x represent the number of games

And let y represent the number of rides

Each game costs $1.25

Each ride costs $2.75

The total expenses incurred by Wyatt amounted to $20.25

Thus, the equation becomes: x1.25 + y2.75 = 20.25..... Equation 1

The number of rides he enjoyed is double the games he competed in

Y = 2x... Equation 2

By substituting the value of y into equation 1

x1.25 + y2.75 = 20.25

x1.25 + 2(x)2.75 = 20.25

x1.25 + x5.5 = 20.25

x6.75= 20.25

x= 20.25/6.75

X= 3

Y= 2x

Y= 2(3)

Y= 6

Thus, Wyatt participated in 3 games and enjoyed 6 rides

5 0

11 days ago

The gravity on the surface of the moon is 0.17 of the surface gravity on Earth. If Jack weighs 156 pounds on Earth, what would b

tester [8842]

Response:

Jack's weight on the moon will be 26.52 pounds

Step-by-step explanation:

x = 156 (Weight on Earth)

y = 0.17 (Moon's gravity)

z = Your weight on the moon

x * y = z

156 * 0.17 = 26.52

Jack will weigh 26.52 pounds.

4 0

11 days ago