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Nitella [24]
2 years ago
10

Use the properties of 30-60-90 and 45-45-90 triangles to solve for x in each of the problems below. Then decode the secret messa

ge by matching the answer with the corresponding letter/symbol from the exercises.

Mathematics
1 answer:
diamong [38]2 years ago
3 0

These problems are solved using the trigonometric function. Trigonometric functions provides the ratio of different sides of a right-angle triangle.

<h3>What are Trigonometric functions?</h3>

The trigonometric function refer to function that are periodic in nature and which lend insight to the relationship between angles and the sides of a triangle that is right angled.

The solutions to x in the respective problems is given as follows:

1st.) x = 5 /Sin(30°)

x = 10

!) sin(45°) = 4/x

x = 4/sin(45°)

x = 4√2

I) Cos(45°) = √3 / x

x = √3 / Cos(45°)

x = √6

E) Tan(60°)

= (3√3) / x

x = (3√3) / 3

W) It is to be noted that for right-triangle that is isosceles in nature, the angle made by the legs and the hypotenuse is always 45°.

x = 45°

N) x² + x² = (7√2)²

x = 7

V) Tan(60°) = 7 / x

x = 7√3/3

K) x² + x² = (9)²

x = 9/√2

Y) Sin(60°) = 7√3/x

x = 14

M) Sin(30°) = x/11

x = 11/2

T) Sin(45°) = x/√10

x = √5

A) x + 2x + 90° = 180°

x = 30°

O) Sin(45°) = √2 / x

x = 2

R) Tan(30°) = x / 4

x = 4/√3

= 4√3 / 3

S) Sin(60°) = x / (10/3)

x = (5√3) / 3

Learn more about Trigonometric functions at:
brainly.com/question/1143565
#SPJ1

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svetoff [14.1K]
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Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

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