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2 years ago

Please help I need help by the end of today

Mathematics

1 answer:

yulyashka [42]2 years ago

6 0

Answer:

A. x = 7, y = 3

Step-by-step explanation:

The triangles are congruent by HL if the hypotenuses are congruent, and if the marked legs are congruent.

3x -1 = 6y +2 . . . . hypotenuses are congruent

2x = 5y -1 . . . . . . .legs are congruent

Subtracting 2 times the first equation from 3 times the second gives ...

3(2x) -2(3x -1) = 3(5y -1) -2(6y +2)

6x -6x +2 = 15y -3 -12y -4

9 = 3y . . . . . . . . . . . add 7 and simplify

y = 3 . . . . . . . . divide by 3

This value of y matches only answer choice A.

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Please help! I've been working on this for a few days and I just don't understand, it's due in a few hours. Thank you.

never [62]

Answer:

Part A: α = arc tan (y/x) = tan⁻¹ (y/x)

Part B: quadrant II α = arc tan (- y/x) = arc (- tan (y/x)) = - tan⁻¹ (y/x) 180°>α>90°

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Step-by-step explanation:

PART A:

In this case we will use trigonometric function tanα to calculate angle α:

tanα = y/x => α = arc tan (y/x) = tan⁻¹ (y/x)

This formula is use in general way and in first quadrant 90°>α>0°

Part B:

But in the other quadrants you must know to use unit circle to reduce angle from II, III and IV quadrant to the first quadrant.

If angle is in the quadrant II 180°>α>90° then

tanα = - tan (180°-α)

If angle is in the quadrant III 270°>α>180° then

tanα = tan (α-180°)

If angle is in the quadrant IV 360°>α>270° then

tanα =- tan (360°-α)

Part C:

vector w (x,y) = (-1,6) this vector lies in quadrant II 180°>α>90°

180°- α = tan⁻¹ (-6/1) = - tan⁻¹ 6 = 80.53° => 180° - α = 80.53°

α = 180° - 80.53 = 99.47°

It's not easy to understand this, but it's not easy for me to explain.

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