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Olegator [25]
3 years ago
15

Describe three ways to determine the measure of segment YZ.

Mathematics
1 answer:
IgorLugansk [536]3 years ago
7 0

Using 3 different methods, we can find that YZ = 25m

For a given triangle rectangle with a known angle θ and a hypotenuse H, we can write the trigonometric relations

sin(θ) = (opposite cathetus)/H

cos(θ) = (adjacent cathetus)/H

tan(θ) = (opposite cathetus)/(adjacent cathetus)

Now, in the given image we can see that:

θ = 30°

H = 50m

XY = adjacent cathetus

YZ = opposite cathetus.

We want to find YZ, so with the known things, we can use the first relation:

sin(30\°) = YZ/50m

sin(30\°)*50m = YZ = 25m

Now let's try another way.

We know that the sum of the internal angles of a triangle is always equal to 180°

In a triangle rectangle, we always have an angle equal to 90°.

Then the missing angle for our triangle can be computed from:

Z + 90° + 30° = 180°

Z = 180° - 90° - 30° = 60°

From this angle, the side YZ is the adjacent cathetus, then we can use the second trigonometric relation:

cos(60\°) = YZ/50m\\cos(60\°)*50m = YZ = 25m\\

Third way.

Whit the same angle we could find the side XY, the opposite cathetus, with:

sin(60\°) = XY/50m \\sin(60\°)*50m = XY = 43.3m

Now that we know one of the catheti, we can use the last trigonometric relation

tan(60\°) = XY/YZ = 43.3m/YZ \\YZ = 43.3m/tan(60\°) = 25m\\

If you want to learn more about triangle rectangles, you can read:

brainly.com/question/24330420

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Answer:

A - f(t)=80(1.5)^t

B - Graph given below

C - Number of trouts in 5th week are 607.2

D - Population of trouts will exceed 500 on the 5th week.

Step-by-step explanation:

We are given that,

The number of trout increases by a factor of 1.5 each week and the initial population of the trout is observed to be 80.

Part A: So, the explicit formula representing the situation is,

f(t)=80(1.5)^t, where f(t) represents the population of trouts after 't' weeks.

Part B: The graph of the function can be seen below.

It can be seen that the function is an exponential function.

Part C: It is required to find the number of trouts in the 5th week.

So, we have,

f(5)=80(1.5)^5

i.e. f(5)=80\times 7.59

i.e. f(5) = 607.2

Thus, the number of trouts in 5th week are 607.2

Part D: We are given that the trout population exceeds 500.

It is required to find the week in which this happens.

So, we have,

50

i.e. (1.5)^t>\frac{500}{80}

i.e. (1.5)^t>6.25

i.e. t\log 1.5>\log 6.25

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i.e. t>\frac{0.7959}{0.1761}

i.e. t > 4.5

As, t represents the number of weeks. So, to nearest whole, t = 5.

Thus, the population of trouts will exceed 500 on the 5th week.

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