PLEASE HELP NEED DONE FAST!!!

How much does the atmosphere weigh over each square meter of surface? express this in tonnes (units of 1000 kg). recall the conv

Anettt [7]

Atmospheric pressure = 101,325 Newtons / m²
Weight per meter square = 101,325 Newtons
Mass per meter square = 101,325 / (9.81 x 1000)
Mass per meter square = 10.3 tonnes per m²

7 0

3 years ago

Find the solutions to the equation below 6x^2+5-4=0

Kipish [7]

Answer:i√6/6 , -i√6/6

Step-by-step explanation:

6x^2+5-4=0

6x^2=-5+4

x=√-1/6

x=±i√1/60=±i√1/√6*(√6/√6)=±i√6/6

3 0

3 years ago

I need to know the opposite of 8 1/4

masya89 [10]

Well Hello!

I believe the answer to your question would be -8 1/4

4 0

3 years ago

Read 2 more answers

A manufactured product has a length that is normally distributed with a mean of 12 cm. The product will be unusable if the lengt

Sav [38]

Answer:

a) σ = 0,1612

b) P [ 11,75 < X < 12,35] = 0,92,44 or 92,44 %

Step-by-step explanation:

NOTE: I assume the unusable length s 11,5 and less

a) For a normal distribution, the probability of P = 0,01 corresponds to z(score) = -3,1 then:

- 3,1 = ( X - μ₀ ) / σ

- 3,1 = (11,5 - 12 )/ σ

-3,1 = - 0,5 / σ

σ = 0,1612

b) If σ = 0,1612

P [ 11,75 < X < 12,35]

z₁ (score) = ( 11,75 - 12 ) / 0,1612

z₁ = - 0,25/ 0,1612

z₁ = -1,55

From z table P [ 11,75 < X] = 0,0606

z₂ (score) = ( 12,35 - 12) / 0,1612

z₂ = 0,35 / 01612

z₂ = 2,17

From z table P [ X < 12,35] = 0,9850

Finally P [ 11,75 < X < 12,35] = 0,9850 - 0,0606

P [ 11,75 < X < 12,35] = 0,92,44 or 92,44 %

5 0

3 years ago

Write the given second order equation as its equivalent system of first order equations. u′′−5u′−4u=1.5sin(3t),u(1)=1,u′(1)=2.5

lakkis [162]

Answer:

hi your question options is not available but attached to the answer is a complete question with the question options that you seek answer to

Answer: v = 5v + 4u + 1.5sin(3t),

0

1

4

5

0

1.5sin(3t)

1

2.5

Step-by-step explanation:

u" - 5u' - 4u = 1.5sin(3t) where u'(1) = 2.5 u(1) = 1

v represents the "velocity function" i.e v = u'(t)

As v = u'(t)

<em>u' = v</em>

since <em>u' = v </em>

v' = u"

v' = 5u' + 4u + 1.5sin(3t) ( given that u" - 5u' - 4u = 1.5sin(3t) )

= 5v + 4u + 1.5sin(3t) ( noting that v = u' )

so v' = 5v + 4u + 1.5sin(3t)

d/dt $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$ = $\left[\begin{array}{ccc}0&1&\\4&5&\\\end{array}\right]$ $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$ + $\left[\begin{array}{ccc}0&\\1.5sin(3t)&\\\end{array}\right]$

Given that u(1) = 1 and u'(1) = 2.5

since v = u'

v(1) = 2.5

note: the initial value for the vector valued function is given as

$\left[\begin{array}{ccc}u(1)&\\v(1)\\\end{array}\right]$ = $\left[\begin{array}{ccc}1\\2.5\\\end{array}\right]$

7 0

3 years ago