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2 years ago

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PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1111111111

1 answer:

Bond [772]2 years ago

4 0

Answer:

261

Step-by-step explanation:

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Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π). (Enter your answers as a comma-separated list

lara31 [8.8K]

Sin(2θ)+sin(<span>θ)=0
use double angle formula: sin(2</span>θ)=2sin(θ)cos(<span>θ).
=>
2sin(</span>θ)cos(θ)+sin(<span>θ)=0
factor out sin(</span><span>θ)
sin(</span>θ)(2cos(<span>θ)+1)=0
by the zero product property,
sin(</span>θ)=0 ...........(a) or
(2cos(<span>θ)+1)=0.....(b)
Solution to (a): </span>θ=k(π<span>)
solution to (b): </span>θ=(2k+1)(π)+/-(π<span>)/3
for k=integer

For [0,2</span>π<span>), this translates to:
{0, 2</span>π/3,π,4π/3}

8 0

2 years ago

What property is illustrated by 5*6=6*5

lara [203]

The commutative property

6 0

2 years ago

Read 2 more answers

Angle 1 and angle 2 form a linear pair. If the measure of angle 1 = (5x+9)º and

True [87]

- linear pairs are supplementary, meaning they add up to 180°

5x + 9 + 3x + 11 = 180

8x + 20 = 180

8x = 160

x = 20

- now plug in.

angle 1: 5(20) + 9 = 100 + 9 = 109°

angle 2: 3(20) + 11 = 60 + 11 = 71°

8 0

2 years ago

If y= 2x + 7 were changed to y = 5x+7, how would the graph of the new

Amanda [17]

Answer:

C is correct

Step-by-step explanation:

Comparing the to equation with slope intercept form: $y=mx+c$

The second equation has the greatest slope, which is indicating that the line is steeper.

4 0

3 years ago

E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis .Find the coordinates of H and G. Please need answe

timurjin [86]

Answer:

coordinates of H = (1, 0)

Coordinates of G = ( - 3.6, -2) or (5.6, -2)

Step-by-step explanation:

E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis.

Let the coordinates of H is (x, 0) and G is (a, b).

The length of side EF is

$EF = \sqrt{(5 -2)^2 + (3 +1)^2} = 5$

So,

$EH = \sqrt{(5 -x)^2 + (3 -0)^2} = 5\\\\(5 -x)^2+ 9 = 25\\\\5 - x = 4\\\\x = 1$

And

$GH = \sqrt{(a -x)^2+ b^2} = 5\\\\(a -1)^2+ b^2 = 25\\\\a^2 + b^2 + 1 - 2 a = 25\\\\a^2 + b^2 - 2a = 24 .... (1)$

Now

$FG = \sqrt{(a -2)^2+ (b +1)^2} = 5\\\\(a -2)^2+ (b+1)^2 = 25\\\\a^2 + b^2 + 2b - 2 a = 20\\ ..... (2)$

Solving (1) and (2)

b = - 2 ,

$a = \frac{2 \pm\sqrt{4 + 80}}{2}\\\\a = \frac{2 \pm 9.2}{2}\\\\a = - 3.6, 5.6$

3 0

2 years ago