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3 years ago

7

the rocks near the shore between two lighthouses at point a and b make the waters unsafe. The measure of AXB is 300. Waters insi

de this arc are unsafe. Suppose you are a navigator on a ship at sea. How can you use the lighthouses to keep the ship in safe waters?

1 answer:

exis [7]3 years ago

6 0

You can, use the lighthouses to know where you are going in the dark night.

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Converting Explicit formula to Recursive formula.

Kruka [31]

The formula
a(n) = 2 - 5(n-1)
is in the form
a(n) = a1 + d(n-1)

where
a1 = first term = 2
d = -5 = common difference

The first term is carried over to the recursive formula. We start with a1 = 2. The next term after that is found by subtracting 5 from the previous term. So
second term = (first term) - 5
third term = (second term) - 5
and so on

The recursive step would be
a(n) = a(n-1)-5

So that's why the answer is choice C

8 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in <em>c)</em>

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

I need help with this simplification-

astra-53 [7]

Answer:

1.31

Step-by-step explanation:

= $\frac{64}{56}$ ^2 x $\frac{1}{2.92}$ ^3 x 2.48

= 1.31 x 0.40 x 2.48

= 0.524 x 2.48

= 1.31

:)

Answer was verified and rounded

7 0

3 years ago

In ΔABC,AB = 20 cm, AC = 15 cm. The length of the altitude AN is 12 cm. Prove that ΔABC is a right triangle.

Makovka662 [10]

It is the pythagorean theorem converse

4 0

3 years ago

Read 2 more answers

the FCMS basketball team has 21 wins and 14 losses write the ratio of wins to total in simplest form. Answer Choices: 21/14 21:3

bogdanovich [222]

Answer:

d. 3/2

Step-by-step explanation:

3 0

3 years ago