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3 years ago

The Seattle Space Needle casts a 67-meter shadow. If the angle of elevation

Mathematics

1 answer:

Alik [6]3 years ago

3 0

Answer:

The space needle is 62.95 m tall.

Step-by-step explanation:

Given that,

The height of the shadow, h = 67 m

The angle of elevation from the tip of the shadow to the top of the Space Needle is 70°.

We need to find the height of the space needle.

The shadow of space needle is hypotenuse of right angled triangle. Let the height of the space needle is x m. Using trigonometry,

$\sin\theta=\dfrac{x}{h}\\\\x=h\times \sin\theta\\\\x=67\times \sin(70)\\\\x=62.95\ m$

So, the space needle is 62.95 m tall.

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Andre has a summer job selling magazine subscriptions. He earns $25 each week plus $3 for every subscription he sells. Andre hop

Strike441 [17]

Answer: 25 + 3n

Step-by-step explanation:

Hi, the answer is lacking the last part:

Write an expression for the amount of money he makes this week.

So, to answer this we have to write an expression:

The fixed amount that he earns per week (25) plus the product of the amount he earns per subscription (3) and the number of subscriptions sold (n) , must be equal to his weekly earnings.

Mathematically speaking:

25 + 3n

Feel free to ask for more if needed or if you did not understand something.

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3 years ago

suppose that the x-intercepts of the graph of y=f(x) are 4 and 8. What are the x-intercepts of the graph of y=f(x+5)

lesya692 [45]

Find what the 5 is and times the f with x

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3 years ago

Two ships leave the same port at the same time. Two hours later ship A has traveled 12 miles and ship B has traveled 8 miles.

sashaice [31]

Answer:

the angle between their paths is 100.8°

Step-by-step explanation:

From the given information, you can construct a triangle, just like the one in the figure.

We will use the Cosine Rule which is:

c² = b² + a² - 2 b c cos(θ)

where

c = 16 miles
b = 8 miles
a = 12 miles

Therefore,

2 b c cos(θ) = b² + a² - c²

cos(θ) = (b² + a² - c²) / 2 b c

θ = cos⁻¹( (b² + a² - c²) / (2 b c) )

θ = cos⁻¹( (8² + 12² - 16²) / 2(8)(16) )

θ = 100.8°

Therefore, the angle between their paths is 100.8°

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3 years ago

Mike has $25.00 to rent paddleboards for himself and a friend for 3 hours. Each paddleboard rental costs $3.75 per hour

zloy xaker [14]

$3.75 x 6 =22.50 so Mike has just enough to rent paddleboards for him and his friend for 3 hours with a total remaining of $2.50

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4 years ago

Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?

irakobra [83]

First, note that $m\angle A+m\angle C=90^{\circ}.$ Then

$m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.$

Consider all options:

$\tan A=\dfrac{\sin A}{\sin C}$

By the definition,

$\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.$

Now

$\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.$

Option A is true.

$\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.$