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Hunter-Best [27]
3 years ago
11

In the figure shown, line AB is parallel to line CD.

Mathematics
1 answer:
disa [49]3 years ago
4 0

Answer:

Part A: <em>x = </em>55°

Step-by-step explanation:

<em>Part A Work:</em>

x + 65 = 120

<em>Part B Work:</em>

<em> </em>The transversal passes through the 2 conforming lines which allows you to use rules such as opposing interior angles. <em> </em>

<em> </em>

<em>hope this helps!</em>

<em>- Kiniwih426</em>

<em />

<em />

<em />

<em />

<em> </em>

<em />

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The coordinates of the vertices of ∆PQR are P(-2,5), Q(-1,1), and R(7,3). Determine whether ∆PQR is a right triangle. Show your
Mashutka [201]

Given

∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

Determine whether ∆PQR is a right triangle

To proof

As given ∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

First find out the sides of triangle

FORMULA

Distance formula between two points

D^{2}= (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}

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PR = \sqrt{(-1+2)^{2}+(1-5)^{2}  }

PR = \sqrt{17}

Distance between two points Q(-1,1)and  R(7,3)

QR = \sqrt{(7+1)^{2} +(3-1)^{2}  }

QR =\sqrt{68}

Distance between two points  R(7,3) and P(-2,5)

RP =\sqrt{(-2-7)^{2} + (5-3)^{2}  }

RP=\sqrt{85}

now show that ∆PQR is a right triangle

RP^{2} = PQ^{2} +QR^{2}

Putting the value given above

(\sqrt{85}) ^{2} = \sqrt{17} ^{2} +\sqrt{68} ^{2}

85 = 17 +68

85 =85

In the right triangle

HYPOTENUSE² = BASE² + PERPENDICULAR²

This is prove above

Hence ∆PQR is a right triangle

Hence proved










7 0
4 years ago
Write an equation that represents the line. (the points) (0,-5) (4,1)
snow_tiger [21]

Answer:

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Step-by-step explanation:

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Please help question 4
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Answer:

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Step-by-step explanation:

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k = \frac{d}{t} = \frac{75}{3} = 25

d = 25t ← equation of variation

when t = 1.6, then

d = 25 × 1.6 = 40 miles

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3 years ago
Given that cos theta = -12/13 and theta is in the second quadrant, find this function. sin theta=​
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5/13

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Answer:

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