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Ksju [112]
3 years ago
15

The point (x, square root of 3/2) is on the unit circle, what is x?

Mathematics
2 answers:
Andreyy893 years ago
5 0

Answer:

x=\frac{1}{2}

Step-by-step explanation:

When we have a point (a,b) on the unit circle, we can say that

a^2+b^2=1

<em>This is a property of the unit circle.</em>

<em />

From the point given  (x,\frac{\sqrt{3} }{2}) , now we can write the equation shown below and solve for x:

x^2+(\frac{\sqrt{3} }{2})^2=1\\x^2+\frac{3}{4}=1\\x^2=1-\frac{3}{4}\\x^2=\frac{1}{4}\\x=\frac{\sqrt{1}}{\sqrt{4} } \\x=\frac{1}{2}

So, x = 1/2

Inessa05 [86]3 years ago
4 0

Answer: x = 1/2

Step-by-step explanation:

We have that the point (x, (√3)/2)) is on the unit circle.

we can define a circle of radius R centered in the (0,0) as:

x^2 + y^2 = R^2

This means that:

x^2 + (√(3)/2)^2 = 1

x^2 + 3/4 = 1

x^2 = 1 - 3/4 = 1/4

x = √(1/4) = 1/2

So we have that x is equal to 1/2

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Consider the equation below.
Korvikt [17]

Answer:

Equation in square form:

y=3(x+5)^2-4

Extreme value:

(h,k)=(-5,-4)

Step-by-step explanation:

We are given

y=3x^2+30x+71

we can complete square

y=3(x^2+10x)+71

we can use formula

a^2+2ab+b^2=(a+b)^2

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now, we can add and subtract 5^2

y=3(x^2+2\times x\times 5+5^2-5^2)+71

y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71

y=3(x+5)^2-75+71

So, we get equation as

y=3(x+5)^2-4

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we know that this parabola

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so, we can compare it with

y=a(x-h)^2+k

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now, we can compare and find h and k

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we get

h=-5

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so, extreme value of this equation is

(h,k)=(-5,-4)

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4 years ago
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