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3 years ago

9

Give one positive and one negative coterminal angle for 135°

1 answer:

8_murik_8 [283]3 years ago

6 0

Answer:

495°, -225°

Step-by-step explanation:

You add and subtract 360° to 135° to get your positve and negative coterminal angle.

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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used

gladu [14]

Answer:

Part 1) $2x^{2}-32=0----> (-4,4)$

Part 2) $4x^{2}-100=0----> (-5,5)$

Part 3) $x^{2}-55=9----> (-8,8)$

Part 4) $x^{2}-140=-19----> (-11,11)$

Part 5) $2x^{2}-18=0----> (-3,3)$

see the attached figure

Step-by-step explanation:

Solve each quadratic equation

Part 1) we have

$2x^{2}-32=0$

Adds 32 both sides

$2x^{2}=32$

Divide by 2 both sides

$x^{2}=16$

take square root both sides

$x=\pm4$

therefore

The solutions are (-4,4)

Part 2) we have

$4x^{2}-100=0$

Adds 100 both sides

$4x^{2}=100$

Divide by 4 both sides

$x^{2}=25$

take square root both sides

$x=\pm5$

therefore

The solutions are (-5,5)

Part 3) we have

$x^{2}-55=9$

Adds 55 both sides

$x^{2}=9+55$

$x^{2}=64$

take square root both sides

$x=\pm8$

therefore

The solutions are (-8,8)

Part 4) we have

$x^{2}-140=-19$

Adds 140 both sides

$x^{2}=-19+140$

$x^{2}=121$

take square root both sides

$x=\pm11$

therefore

The solutions are (-11,11)

Part 5) we have

$2x^{2}-18=0$

Adds 18 both sides

$2x^{2}=18$

Divide by 2 both sides

$x^{2}=9$

take square root both sides

$x=\pm3$

therefore

The solutions are (-3,3)

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Step-by-step explanation:

I hope its right

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