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3 years ago

Find the two smallest possible solutions to part 1a

Mathematics

2 answers:

bixtya [17]3 years ago

7 0

<h3>Answer: A. 5/12, 25/12</h3>

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Work Shown:

12*sin(2pi/5*x)+10 = 16

12*sin(2pi/5*x) = 16-10

12*sin(2pi/5*x) = 6

sin(2pi/5*x) = 6/12

sin(2pi/5*x) = 0.5

2pi/5*x = arcsin(0.5)

2pi/5*x = pi/6+2pi*n or 2pi/5*x = 5pi/6+2pi*n

2/5*x = 1/6+2*n or 2/5*x = 5/6+2*n

x = (5/2)*(1/6+2*n) or x = (5/2)*(5/6+2*n)

x = 5/12+5n or x = 25/12+5n

these equations form the set of all solutions. The n is any integer.

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The two smallest positive solutions occur when n = 0, so,

x = 5/12+5n or x = 25/12+5n

x = 5/12+5*0 or x = 25/12+5*0

x = 5/12 or x = 25/12

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Plugging either x value into the expression 12*sin(2pi/5*x)+10 should yield 16, which would confirm the two answers.

Mkey [24]3 years ago

4 0

Answer:

Step-by-step explanation:

12sin(⅖pi x) + 10 = 16

sin(⅖pi x) = 6/12

⅖pi x = sin^-1(½)

⅖pi x = pi/6, pi - pi/6

⅖pi x = pi/6, 5pi/6

x = 5/12, 25/12

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3 years ago

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allsm [11]

Answer:

The answer is number (2) ⇒ because of the AAS congruence theorem

Step-by-step explanation:

* Lets use the information to solve the problem

- In the given triangles TSR and QRS

# We have a common side RS or SR

# Two right angles TSR and QRS

# m∠T = m∠Q ⇒ given

* So we have two pairs of angles and one common side, lets

revise the cases of congruence

- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ

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including angle in the 2nd Δ

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ

≅ 2 angles and the side whose joining them in the 2nd Δ

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles

and one side in the 2ndΔ

* Now lets read the statements and write the missing

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Step 2: m∠T = m∠Q ⇒ because it is given

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3 years ago

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mr_godi [17]

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2 years ago

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Harrizon [31]

The correct question is:

Suppose x = c1e^(-t) + c2e^(3t) a solution to x''- 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)

Answer:

x = c1e^(-t) + c2e^(3t)

is a solution to the differential equation

x''- 2x' - 3x = 0

Step-by-step explanation:

We need to verify that

x = c1e^(-t) + c2e^(3t)

is a solution to the differential equation

x''- 2x' - 3x = 0

We differentiate

x = c1e^(-t) + c2e^(3t)

twice in succession, and substitute the values of x, x', and x'' into the differential equation

x''- 2x' - 3x = 0

and see if it is satisfied.

Let us do that.

x = c1e^(-t) + c2e^(3t)

x' = -c1e^(-t) + 3c2e^(3t)

x'' = c1e^(-t) + 9c2e^(3t)

Now,

x''- 2x' - 3x = [c1e^(-t) + 9c2e^(3t)] - 2[-c1e^(-t) + 3c2e^(3t)] - 3[c1e^(-t) + c2e^(3t)]

= (1 + 2 - 3)c1e^(-t) + (9 - 6 - 3)c2e^(3t)

= 0

Therefore, the differential equation is satisfied, and hence, x is a solution.

4 0

3 years ago

Find the two smallest possible solutions to part 1a​

Find the two smallest possible solutions to part 1a