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3 years ago

I swear to god this is urgent:

Mathematics

1 answer:

Illusion [34]3 years ago

6 0

the line-of-sight distance from the television camera to the base of the stadium is 1449.28 m .

Step-by-step explanation:

A blimp provides aerial television views of a baseball game. The television camera sights the stadium at a 12° angle of depression. The altitude of the blimp is 300 m. We need to find What is the line-of-sight distance from the television camera to the base of the stadium . Let's find out:

According to question , given scenario is in a right angle triangle where

$Perpendicular=300\\Hypotenuse=?\\x=12$ , where x is angle of depression.

We know that $sinx= \frac{Perpendicular}{Hypotenuse}$

⇒ $sin12= \frac{300}{Hypotenuse}$

⇒ $Hypotenuse= \frac{300}{Sin12}$

⇒ $Hypotenuse= \frac{300}{0.207}$

⇒ $Hypotenuse= 1449.28m$

Therefore , the line-of-sight distance from the television camera to the base of the stadium is 1449.28 m .

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Answer:

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Step-by-step explanation:

Given that:

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We will find now that 25 is what percent of 75 so:

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3 years ago

Read 2 more answers

A cylinder has radius r and height h. A. How many times greater is the surface area of a cylinder when both dimensions are multi

Ierofanga [76]

Answer: A. Factor 2 => 4x greater

Factor 3 => 9x greater

Factor 5 => 25x greater

Step-by-step explanation: A. A cylinder is formed by 2 circles and a rectangle in the middle. That's why surface area is given by circumference of a circle, which is the length of the rectangle times height of the rectangle, i.e.:

A = 2.π.r.h

A cylinder of radius r and height h has area:

$A_{1}$ = 2πrh

If multiply both dimensions by a factor of 2:

$A_{2}$ = 2.π.2r.2h

$A_{2}$ = 8πrh

Comparing $A_{1}$ to $A_{2}$ :

$\frac{A_{2}}{A_{1}}$ = $\frac{8.\pi.rh}{2.\pi.rh}$ = 4

Doubling radius and height creates a surface area of a cylinder 4 times greater.

By factor 3:

$A_{3} = 2.\pi.3r.3h$

$A_{3} = 18.\pi.r.h$

Comparing areas:

$\frac{A_{3}}{A_{1}}$ = $\frac{18.\pi.r.h}{2.\pi.r.h}$ = 9

Multiplying by 3, gives an area 9 times bigger.

By factor 5:

$A_{5} = 2.\pi.5r.5h$

$A_{5} = 50.\pi.r.h$

Comparing:

$\frac{A_{5}}{A_{1}}$ = $\frac{50.\pi.r.h}{2.\pi.r.h}$ = 25

The new area is 25 times greater.

B. By analysing how many times greater and the factor that the dimensions are multiplied, you can notice the increase in area is factor². For example, when multiplied by a factor of 2, the new area is 4 times greater.

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3 years ago