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oksano4ka [1.4K]
3 years ago
8

a ship is 2.6° off course. if the ship is traveling at 26.0 miles per hour, how far off course will it be after 5 hours? (round

your answer to two decimal places.)
Mathematics
1 answer:
REY [17]3 years ago
3 0
2.6*5=13 degrees after 5 hours at 2.6 degrees per hour


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A storage company must design a large rectangular container with a square base. The volume is 6750ft3. The material for the top
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3 years ago
Help...please I need this fast
Simora [160]

Answer:

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3 0
3 years ago
If m ∠ A B F = ( 8 w − 6 ) ° m ∠ A B E = [ 2 ( w + 11 ) ] ° m ∠ E B F .
pashok25 [27]

The measure of angle EBF where he angle measures are given as m∠ABF = (8w − 6)° and m∠ABE = [2(w + 11)] is m∠EBF = 4w - 28

<h3>How to determine the measure of angle EBF?</h3>

The angle measures are given as

If m ∠ A B F = ( 8 w − 6 ) ° m ∠ A B E = [ 2 ( w + 11 ) ] ° m ∠ E B F

Rewrite the angle measures properly.

This is done, as follows

m∠ABF = (8w − 6)°

m∠ABE = [2(w + 11)]

The measure of angle m∠EBF is calculated as:

m∠ABF = m∠ABE + m∠EBF

Substitute the known values in the above equation

8w - 6 = 2(2w + 11) + m∠EBF

Open the brackets

8w - 6 = 4w + 22 + m∠EBF

Evaluate the like terms

m∠EBF = 4w - 28

Hence, the measure of angle EBF where he angle measures are given as m∠ABF = (8w − 6)° and m∠ABE = [2(w + 11)] is m∠EBF = 4w - 28

Read more about angles at

brainly.com/question/25716982

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5 0
2 years ago
Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A
Vlad [161]

Answer:

y(x)=sin(2x)+2cos(2x)

Step-by-step explanation:

y''+4y=0

This is a homogeneous linear equation. So, assume a solution will be proportional to:

e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda

Now, substitute y(x)=e^{\lambda x} into the differential equation:

\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0

Using the characteristic equation:

\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0

Factor out e^{\lambda x}

e^{\lambda x}(\lambda ^2 +4) =0

Where:

e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda

Therefore the zeros must come from the polynomial:

\lambda^2+4 =0

Solving for \lambda:

\lambda =\pm2i

These roots give the next solutions:

y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}

Where c_1 and c_2 are arbitrary constants. Now, the general solution is the sum of the previous solutions:

y(x)=c_1 e^{2ix} +c_2 e^{-2ix}

Using Euler's identity:

e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)

y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\

Redefine:

i(c_1-c_2)=c_1\\\\c_1+c_2=c_2

Since these are arbitrary constants

y(x)=c_1sin(2x)+c_2cos(2x)

Now, let's find its derivative in order to find c_1 and c_2

y'(x)=2c_1 cos(2x)-2c_2sin(2x)

Evaluating    y(0)=2 :

y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2

Evaluating     y'(0)=2 :

y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1

Finally, the solution is given by:

y(x)=sin(2x)+2cos(2x)

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