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3 years ago

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A man standing at point A walks 1 mile south, 3 miles east, 5 miles north, and 7 miles west to reach point B. The length of line

segment
AB is
A. 4 miles
B. √8 miles
C. 2 miles
D. √32miles
E. √12 miles

1 answer:

joja [24]3 years ago

3 0

Answer: √32

Step-by-step explanation: so the displacement is north 4 and west 4 and it becomes a right triangle so the hypotenuse is the distance for segment AB

So Drawn out should like straight lines up,down,left,and right. So east is positive while west is negative same for North(+) and South(-)

I recommend sticking with 2 directions first. Starting with north and south right. First you go south 1 mile but then you go north 5miles in math terms

-1+5=4 or down one is one less than 5

Same for east and west. 3+-7 = -4

So we went positive 4 miles up and 4miles to the left

So (-4)^2+4^2= hypotenuse or AB segment in this case

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If 10 pencils and 7 erasers cost $4.23 and 3 pencils and 1 eraser cost $0.95, what is the cost of 2 erasers?

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Answer: $0.58

<u>Step-by-step explanation:</u>

Let x represent pencil and y represent eraser

10x + 7y = 4.23 → 1(10x + 7y = 4.23) → 10x + 7y = 4.23

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3 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

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Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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