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Marat540 [252]
2 years ago
13

Please help math homeowke please and Tysm

Mathematics
1 answer:
12345 [234]2 years ago
6 0

The length of the line segment TV is 30.

<h3>How to find the segment length by solving linear equations</h3>

In this problem we have a geometrical system formed by two collinear line segments (TU, UV). Mathematically speaking, the geometrical system is described by the following equations:

TV = TU + UV       (1)

TU = 25                 (2)

UV = 2 · x + 15       (3)

TV = x + 35            (4)

By (1), (2), (3) and (4):

x + 35 = 25 + (2 · x + 15)

x + 35 = 2 · x + 40

35 - 40 = 2 · x - x

- 5 = x

x = - 5

Lastly, we evaluate (4) at x = - 5 and find the length of the line segment TV is:

TV = - 5 + 35

TV = 30

The length of the line segment TV is 30.

To learn more on line segments: brainly.com/question/25727583

#SPJ1

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How is this solved using trig identities (sum/difference)?
GenaCL600 [577]
FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

cos \beta = -\frac{1}{2}  \sqrt{3}

tan \beta= \frac{1}{3}  \sqrt{3}

SECOND PART
Solve the questions
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sin (α + β) = sin α cos β + cos α sin β
sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2}  \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )
sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )
sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )
cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )
cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )

Find tan (α - β)
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tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})
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