#11 PLEASE HELP!!! URGENT!! NEED ASAP!!!

Answer:

9/50 rad/sec

Step-by-step explanation:

let the angle be represented as ∅

distance form bottle rocket on the ground = 20 ft

height of rocket = 40 ft

solution

first we differentiate w.r.t.time

= sec²(∅)*(d(∅)/dt))=(1/20)*dh/dt   --------- ( 1 )

=  d(∅)/dt = (1/20)cos²(∅)* (dh/dt)   ------- ( 2 )

note :  1/sec(∅) =cos(∅)

therefore ; dh/dt = 18 ft/sec

applying Pythagoras theorem

Tan ∅ =  40 / 20

 h^2( hypotenuse )  = a^2( height )  + b^2 ( distance from bottle rocket )

                     h = √ 40^2 + 20^2 =  √ 2000

also cos ( ∅ ) = b / h  =  20 / √ 2000

hence cos²∅ = 1/5

Back to equation 2

d(∅)/dt=(1/20)×(1/5)×(18)

= 18/100  = 9/50 ( changing rate of angle of elevation )

Answer:

yo done so easy

Step-by-step explanation:

Answer:

y-intercept = (0, 8)

axis of symmetry= -b/2a

Step-by-step explanation:

Also when you want something to be squared use this symbol ^

Answer:

45

Step-by-step explanation:

i believe its 45 or 55

Answer:

92°

Step-by-step explanation:

(2x - 2)° = x ° + 45°

(2x - 2 - x)° = 45°

(x - 2)° = 45°

x - 2 = 45

x = 45 + 2

x = 47

(2x - 2)° = (2*47 - 2)° = (94 - 2)* = 92°

Answer:

The 95% confidence interval is   [tex]0.6428 < \sigma < 1.0954[/tex]

Step-by-step explanation:

From the question we are told that

   The sample size is  n = 29

    The sample standard deviation is [tex]s = 0.81[/tex]

  Generally the 95% confidence interval to estimate the true standard deviation  is mathematically represented as  

                      [tex]\sqrt{\frac{(n- 1)s^2 }{ X^2_( \frac{\alpha }{2 } , n-1)} } < \sigma < \sqrt{\frac{(n- 1)s^2 }{ X^2_(1- \frac{\alpha }{2 } , n-1)} }[/tex]

Generally from the  chi -distribution table  the critical value of [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of  [tex]df = 29 - 1 = 28[/tex]  is  

            [tex]X^2 _{\frac{\alpha }{2} , 28 } =44.46079184[/tex]

Generally from the  chi -distribution table  the critical value of [tex]1 - \frac{\alpha }{2}[/tex] at a degree of freedom of  [tex]df = 29 - 1 = 28[/tex]  is  

             [tex]X^2 _{(1 - \frac{\alpha }{2}) , 28 } = 15.30786055[/tex]

So

                 [tex]\sqrt{\frac{(29- 1)0.81^2 }{ 44.46079184 } } < \sigma < \sqrt{\frac{(29- 1)0.81^2 }{15.315.307860551 }[/tex]

=>              [tex]0.6428 < \sigma < 1.0954[/tex]

i would go with either a or b

Answer:

Least to greatest: -5 1/2, -5.2, -5, -5/2, 5.5

Answer:

The student is incorrect because I I or absolute value would make the -9 to positive 9.  

Step-by-step explanation:

-9<4

on a numberline

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4

but the absolute value of -9 (or) |–9|

is actually positive 9

9>4

on a number line

4 5 6 7 8 9

Odpowiedź:

48 cm

Wyjaśnienie krok po kroku:

Jeśli się uwzględni:

Wysokość trapezu = 6 cm

Długość podstaw:

Podstawa 1 = a = 9 cm

Podstawa 2 = b = (a - 2) = 9 - 2 = 7 cm

Uaing relacji:

Powierzchnia trapezu = 1/2 (a + b) * h

Powierzchnia trapezu = 0,5 (9 + 7) * 6

Powierzchnia trapezu = 0,5 (16) * 6

Powierzchnia trapezu = 8 * 6

Powierzchnia trapezu = 48 cm

Answer:

23 + 17 = 40 :)

Answer:

2x+48=6x

Step-by-step explanation:

Answer:

It's the last one

I hoped that helped