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Nuetrik [128]
3 years ago
12

HELP!! PLEASE!! NEED DONE ASAP I WILL GIVE 100 POINTS AND MORE IF YOU CAN GET ALL 3 QUESTIONS DONE!! THANK YOU!! I WILL ALSO MAR

K BRAINLIST Kayson is looking at two buildings, building A and building B, at an angle of elevation of 73°. Building A is 30 feet away, and building B is 35 feet away. Which building is taller and by approximately how many feet? Building A; it is around 16.35 feet taller than building B Building A; it is around 114.48 feet taller than building B Building B; it is around 16.35 feet taller than building A Building B; it is around 114.48 feet taller than building A Look at the figure below: an image of a right triangle is shown with an angle labeled y If sin y° = a divided by 6 and tan y° = a divided by b, what is the value of cos y°? cos y° = 6 divided by b cos y° = 6a cos y° = 6b cos y° = b divided by 6 If sin f° = eight ninths and the measure of segment YW is 24 units, what is the measure of segment YX? triangle XYW in which angle W is a right angle, angle X measure f degrees, and angle Y measures d degrees 21 units 24 units 27 units 30 units

Mathematics
2 answers:
mr Goodwill [35]3 years ago
5 0

Answer:

The third one is the answer

Building B; it is around 16.35 feet taller than building A

Look at the figure below: an image of a right triangle is shown with an angle labeled y If sin y° = a divided by 6 and tan y° = a divided by b, what is the value of cos y°?

cos y° = 6 divided by b cos y° = 6a cos

If sin f° = 8/9 and the measure of YW is 24 units, what is the measure of YX?

24 units

VARVARA [1.3K]3 years ago
4 0

Answer:

I answered the questions but that formatting is very confusing and discourages anyone for trying to answer.

The last question is confusing. If I got it wrong, tell me. I will try to answer in the comment section then.

Step-by-step explanation:

Kayson is looking at two buildings, building A and building B, at an angle of elevation of 73°. Building A is 30 feet away, and building B is 35 feet away. Which building is taller and by approximately how many feet?

$\text{tan}\theta=\frac{h_{A}}{30} \Rightarrow h_{A}=\text{tan}73\º \cdot 30 \Rightarrow h_{A}\approx 98.12 \text{ feet}$

$\text{tan}\theta=\frac{h_{B}}{35} \Rightarrow h_{B}=\text{tan}73\º \cdot 35 \Rightarrow h_{B}\approx 114.47 \text{ feet}$

\text{Difference} \approx 16.35

Building B is around 16.35 feet taller than building A.

A Look at the figure below: an image of a right triangle is shown with an angle labeled y If sin y° = a divided by 6 and tan y° = a divided by b, what is the value of cos y°?

$\text{sin}(y)=\frac{a}{6} $

$\text{tan}(y)=\frac{a}{b} $

a is the opposite side; 6 is the hypotenuse; b is the adjacent side.

Therefore,

$\text{cos}(y)=\frac{b}{6} $

If sin f° = eight ninths and the measure of segment YW is 24 units, what is the measure of segment YX? triangle XYW in which angle W is a right angle, angle X measure f degrees, and angle Y measures d degrees.

This seems a bit confusing. The angles don't match. We have 90\º+62\º +21\º \approx 173\º

\text{sin}(f)=0.888...

f \approx 62\º

YX is the hypotenuse of the right triangle.

$\text{cos}(21\º)=\frac{24}{YX} $

YX \approx 25.7

Considering f \approx 69\º

$\text{sin}(69\º)\approx\frac{24}{YX} $

YX \approx 25.7

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Answer:

\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

Step-by-step explanation:

Given

f(x) = \sin x\\

c = \frac{3\pi}{4}

Required

Find the Taylor series

The Taylor series of a function is defines as:

f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n

We have:

c = \frac{3\pi}{4}

f(x) = \sin x\\

f(c) = \sin(c)

f(c) = \sin(\frac{3\pi}{4})

This gives:

f(c) = \frac{1}{\sqrt 2}

We have:

f(c) = \sin(\frac{3\pi}{4})

Differentiate

f'(c) = \cos(\frac{3\pi}{4})

This gives:

f'(c) = -\frac{1}{\sqrt 2}

We have:

f'(c) = \cos(\frac{3\pi}{4})

Differentiate

f"(c) = -\sin(\frac{3\pi}{4})

This gives:

f"(c) = -\frac{1}{\sqrt 2}

We have:

f"(c) = -\sin(\frac{3\pi}{4})

Differentiate

f"'(c) = -\cos(\frac{3\pi}{4})

This gives:

f"'(c) = - * -\frac{1}{\sqrt 2}

f"'(c) = \frac{1}{\sqrt 2}

So, we have:

f(c) = \frac{1}{\sqrt 2}

f'(c) = -\frac{1}{\sqrt 2}

f"(c) = -\frac{1}{\sqrt 2}

f"'(c) = \frac{1}{\sqrt 2}

f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n

becomes

f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n

Rewrite as:

f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n

Generally, the expression becomes

f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

Hence:

\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

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