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Viktor [21]
3 years ago
6

I’m not sure if it would be A or B, does anyone know?

Mathematics
1 answer:
KengaRu [80]3 years ago
5 0
B.  It is vertically stretched by a factor of 4.  If you had x^2 versus 4x^2, when x is positive and increasing, 4x^2 is increased by 4 for every value of x.
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A jar of jelly beans contains 50 red gumballs<span> , 45 yellow gumballs, and 30 green gumballs. You reach into the jar and randomly select a jelly bean, then select another while replacing  the first jelly bean back. What is the probability that you draw two red jelly beans? This is Independent because you put the other jelly bean in thus keeping the total number of jelly beans.</span>

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2 years ago
For 0 ≤ θ &lt; 2 π what are the solutions to sin^2(θ) =2sin^2(θ/2)
gregori [183]

Recall the half-angle identity for sine,

sin²(x/2) = (1 - cos(x))/2

Then the given equation is identical to

sin²(θ) = 1 - cos(θ)

Also recall the Pythagorean identity,

sin²(θ) + cos²(θ) = 1

Then we rewrite the equation as

1 - cos²(θ) = 1 - cos(θ)

Factoring the left side, we have

(1 - cos(θ)) (1 + cos(θ)) = 1 - cos(θ)

and so

(1 - cos(θ)) (1 + cos(θ)) - (1 - cos(θ)) = 0

and we factor this further as

(1 - cos(θ)) (1 + cos(θ) - 1) = 0

which gives

cos(θ) (1 - cos(θ)) = 0

Then either

cos(θ) = 0   or   1 - cos(θ) = 0

cos(θ) = 0   or   cos(θ) = 1

[θ = arccos(0) + 2nπ   or   θ = -arccos(0) + 2nπ]

…   or   [θ = arccos(1) + 2nπ   or   θ = -arccos(1) + 2nπ]

(where n is any integer)

[θ = π/2 + 2nπ   or   θ = -π/2 + 2nπ]   or   [θ = 0 + 2nπ]

In the interval 0 ≤ θ < 2π, we get three solutions:

• first solution set with n = 0   ⇒   θ = π/2

• second solution set with n = 1   ⇒   θ = 3π/2

• third solution set with n = 0   ⇒   θ = 0

So, the first choice is correct.

6 0
3 years ago
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