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frutty [35]
3 years ago
13

Given: cos θ=-4/5, sin x = -12/13, θ is in the third quadrant, 

Mathematics
1 answer:
USPshnik [31]3 years ago
4 0

By definition of tangent,

tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)

Recall the double angle identities:

sin(2<em>θ</em>) = 2 sin(<em>θ</em>) cos(<em>θ</em>)

cos(2<em>θ</em>) = cos²(<em>θ</em>) - sin²(<em>θ</em>) = 2 cos²(<em>θ</em>) - 1

where the latter equality follows from the Pythagorean identity, cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1. From this identity we can solve for the unknown value of sin(<em>θ</em>):

sin(<em>θ</em>) = ± √(1 - cos²(<em>θ</em>))

and the sign of sin(<em>θ</em>) is determined by the quadrant in which the angle terminates.

<em />

We're given that <em>θ</em> belongs to the third quadrant, for which both sin(<em>θ</em>) and cos(<em>θ</em>) are negative. So if cos(<em>θ</em>) = -4/5, we get

sin(<em>θ</em>) = - √(1 - (-4/5)²) = -3/5

Then

tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)

tan(2<em>θ</em>) = (2 sin(<em>θ</em>) cos(<em>θ</em>)) / (2 cos²(<em>θ</em>) - 1)

tan(2<em>θ</em>) = (2 (-3/5) (-4/5)) / (2 (-4/5)² - 1)

tan(2<em>θ</em>) = 24/7

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Shkiper50 [21]

<u>w</u><u>=</u><u>-</u><u>1</u><u>.</u><u>5</u>

<u>w</u><u>=</u><u>2</u>

Answer:

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\sqrt{2w²-19w+31}+2=7-2w

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\sqrt{2w²-19w+31}=7-2w-2

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Squaring on both side

(\sqrt{2w²-19w+31})²=(5-2w)²

2w²-19w+31=5²-2*5*2w+4w²

take terms one side

2w²-19w+31-25+20w-4w²=0

-2w²+w+6=0

2w²-w-6=0

doing middle term factorisation

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W=-3/2=-1.5

3 0
3 years ago
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A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po
Ronch [10]

<em>Directed numbers</em> are numbers that have either a <u>positive</u> or <u>negative </u>sign, which can be shown on a <em>number line</em>. Therefore, point F is Fifteen-halves of line <em>segment</em> DE.

A <u>number line</u> is a system that can show the positions of <em>positive</em> or <em>negative</em> numbers. It has its <em>ends</em> ranging from <em>negative infinity</em> to <em>positive infinity</em>. Thus any <em>directed</em> number can be located on the line.

Directed numbers are numbers with either a <u>negative</u> or <u>positive </u>sign, which shows their direction with respect to the <em>number line.</em>

In the given question, the <u>distance</u> between points D and E is <em>9 units</em>. So that <em>dividing</em> 9 units in the ratio of 5 to 6, we have;

\frac{5}{6} x 9   = \frac{45}{6}

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Therefore, the <em>location</em> of point F, which <u>partitions</u> the directed line segment from d to E into a 5:6 ratio is    \frac{15}{2}. Thus the<em> answer</em> is <u>Fifteen-halves.</u>

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7 0
2 years ago
The length of a rectangle is 7 more than the width. The area is 744 sqaure yards, find the length and width of the rectangle​
AVprozaik [17]

Answer:

  • Length of rectangle is 31 yards and Width is 24 yards.

Given:

  • The length of a rectangle is 7 more than the width.
  • The area is 744 sqaure yards

Solution:

Let's assume Width of rectangle be x and Length of rectangle be x + 7 respectively.

Using formula

\\  \:  \:  \:  \:  \pink{ \dashrightarrow \:  \:  \:  \:  \sf { \underbrace{Area_{(Rectangle)} =  Length × Width  }}} \\  \\

On Substituting the required values, we get;

\\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf (x)(x + 7) = 744 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 7x = 744 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 7x - 744 = 0 \\ \\ \\  \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 31x - 24x - 744 = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf x(x + 31) - 24 (x + 31) = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf (x + 31)(x - 24) = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf x = 24 \:  or \:   - 31 \\ \\

As we know that width of the rectangle can't be negative. So x = 24

<u>Hence</u>,

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  • Length of the rectangle = x + 7 = 31 yards

\thereforeLength of rectangle is 31 yards<u> </u>and Width is 24 yards.

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Answer:

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Step-by-step explanation:

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5 0
3 years ago
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