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2 years ago

An 80-m tower is supported by a guy wire attached to the top of the tower. If the wire forms an

1 answer:

5 0

The Sine or Sinθ in a right-angle triangle is the ratio of its perpendicular to its Hypotenuse. The length of the wire is 81.5 meters.

The Sine or Sinθ in a right-angle triangle is the ratio of its perpendicular to its Hypotenuse. it is given as,

Sin(θ) = Perpendicular/Hypotenuse

where,

θ is the angle,

Perpendicular is the side of the triangle opposite to the angle θ,

The hypotenuse is the longest side of the triangle.

The length of the tower is 80 meters, while the angle of elevation is 79°. Therefore, the length of the wire will be the hypotenuse of the triangle. Therefore, the length of the wire is,

Sin(θ) = Perpendicular/Hypotenuse

Sin(79°) = 80 meter/Length of the wire

Length of the wire = 81.4973 ≈ 81.5 meters

Hence, the length of the wire is 81.5 meters.

Learn more about Sine:

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If we wanted to measure the capacity of the waterfall and how deep it was, what unit of measure would we use?

AleksAgata [21]

Probably dept and lenght

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3 years ago

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

3 years ago

I need help with slope (8th grade)

Triss [41]

Answer:

(y1-y2)/(x1-x2)

Step-by-step explanation:

if you have two points from the graph then you can use them to find the slope, for example:

(2,3) and (0,5)

you would take the first point and put it in a fraction (and the second point) but the x goes on the bottom and the y goes on the top:

3/2 and 5/0

Next you put them together minusing one of the equations, but make sure that the two coordinates line up (it doesn't matter which one, in the order):

3-5/2-0

Then you solve:

-2/2

-1

which means that -1 is the slope for this

(the "/" is a fraction bar, in case you didn't already know that)

5 0

3 years ago

Find the domain of the following rational expression f(x)=4x-8/5

Alchen [17]

4x is the correct answer

7 0

3 years ago

ERROR ANALYSIS Describe and correct the error in

valkas [14]

Answer:

The graph should be stretched rather than become narrower.

Step-by-step explanation:

To figure this out, just create some example points.

At x = 0, your y-value will always be 0. But if you were to plug in the value 1, you would get a y-value of 1 in y=x^2, but a value of 0.5 in y=0.5x^2. If you were to plug in a value of 2, you would get a value of 4 in y=x^2, but a value of 2 in y=0.5x^2.

If you continue this pattern for a few more points, then plot them, you will see that adding a coefficient of 0.5 simply stretches the graph

3 0

2 years ago