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2 years ago

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Triangle WXY, with a vertex X at (3, 0), is rotated clockwise 270° about the origin. Which of the following rotations is equival

ent 270° clockwise rotation about the origin?

2 answers:

sergey [27]2 years ago

8 0

What are the answer options?

A rotation of 270 would be equal to 630, 990, etc. Adding 360 degrees would result in the same rotation.

A rotation of 90 degrees counter-clockwise would also result in the same rotation, as would 90 plus any multiple of 360 if rotated counter-clockwise.

Nezavi [6.7K]2 years ago

5 0

The answer is b.

Step-by-step explanation:

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(i) 3 (a + 5b) + a (a + 4) (ii) y (10 - y) + 3 (y - 2) (iii) 2 (8a - 5b) + 3 (5a - 12) (iv) 3 (y - 3) + ( 8 - 6y + x) (v) a (a -

AveGali [126]

Answer:

$3 (a + 5b) + a (a + 4) = a^2 + 7a + 15b$

$y (10 - y) + 3 (y - 2) = - y^2+13y - 6$

$2 (8a - 5b) + 3 (5a - 12) = 31a -10b - 36$

$3 (y - 3) + ( 8 - 6y + x) = -3y + x-1$

$a (a - 2b) + b (b + 2a - c) =a^2 + b^2 - bc$

$5 (x - y + z) + (4x + 3y) =9x - 2y +5z$

Step-by-step explanation:

Solving (i):

$3 (a + 5b) + a (a + 4)$

Open brackets

$3 (a + 5b) + a (a + 4) = 3a + 15b + a^2 + 4a$

Collect like terms

$3 (a + 5b) + a (a + 4) = a^2 + 4a+3a + 15b$

$3 (a + 5b) + a (a + 4) = a^2 + 7a + 15b$

Solving (ii)

$y (10 - y) + 3 (y - 2)$

Open bracket

$y (10 - y) + 3 (y - 2) = 10y - y^2 + 3y - 6$

Collect like terms

$y (10 - y) + 3 (y - 2) = - y^2+10y + 3y - 6$

$y (10 - y) + 3 (y - 2) = - y^2+13y - 6$

Solving (iii)

$2 (8a - 5b) + 3 (5a - 12)$

Open bracket

$2 (8a - 5b) + 3 (5a - 12) = 16a -10b + 15a - 36$

Collect like terms

$2 (8a - 5b) + 3 (5a - 12) = 16a+ 15a -10b - 36$

$2 (8a - 5b) + 3 (5a - 12) = 31a -10b - 36$

Solving (iv)

$3 (y - 3) + ( 8 - 6y + x)$

Open bracket

$3 (y - 3) + ( 8 - 6y + x) = 3y - 9 + 8 - 6y + x$

Collect like terms

$3 (y - 3) + ( 8 - 6y + x) = 3y - 6y + x- 9 + 8$

$3 (y - 3) + ( 8 - 6y + x) = -3y + x-1$

Solving (v):

$a (a - 2b) + b (b + 2a - c)$

Open bracket

$a (a - 2b) + b (b + 2a - c) =a^2 - 2ab + b^2 + 2ab - bc$

Collect like terms

$a (a - 2b) + b (b + 2a - c) =a^2 + 2ab- 2ab + b^2 - bc$

$a (a - 2b) + b (b + 2a - c) =a^2 + b^2 - bc$

Solving (vi)

$5 (x - y + z) + (4x + 3y)$

Open brackets

$5 (x - y + z) + (4x + 3y) =5x - 5y +5z+4x +3y$

Collect like terms

$5 (x - y + z) + (4x + 3y) =5x +4x +3y- 5y +5z$

$5 (x - y + z) + (4x + 3y) =9x - 2y +5z$

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