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musickatia [10]
3 years ago
7

Which of the following explains why cos60 = sin30 using the unit circle?

Mathematics
2 answers:
amid [387]3 years ago
7 0
Hey there

Statement (A) tells us why <span>cos60 = sin30 using the unit circle. 

(A) = </span><span>The side opposite a 30° angle is the same as the side adjacent to a 60° angle in a right triangle. On a unit circle, the y (sin) distance of a 30° angle is the same as the x (cos) distance of a 60° angle.</span>
kaheart [24]3 years ago
7 0
A is the correct answer.  The sine pertains to the opposite side of a right triangle while cosine pertains to the adjacent side.  On the unit circle, x represents cosine and y represents sine.
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Y=6x+-3

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What are the new<br> coordinates of (6,-2) with<br> the dilation scale factor of<br> 1/2.
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(3, - 1 )

Step-by-step explanation:

Assuming the dilatation is centred at the origin then multiply each coordinate by \frac{1}{2}

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In a large on-the-job training program, half of the participants are female and half are male. In a random sample of six partici
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If Angle 8 is congruent to angle 10 and Angle 1 is congruent to angle 7, which describes all the lines that must be parallel? Li
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Multiply 3/sqrt17- sqrt2 by which fraction will produce an equivalent fraction with rational denominator
zzz [600]

Answer:

B.

Step-by-step explanation:

To simplify something that looks like \frac{\text{whatever}}{\sqrt{a}-\sqrt{b}} you would multiply the top and bottom by the conjugate of the bottom. So you multiply the top and bottom for this problem I just made by:

\sqrt{a}+\sqrt{b}.

If you had  \frac{\text{whatever}}{\sqrt{a}+\sqrt{b}}, then you would multiply top and bottom the conjugate of \sqrt{a}+\sqrt{b} which is \sqrt{a}-\sqrt{b}.

The conjugate of a+b is a-b.

These have a term for it because when you multiply them something special happens.  The middle terms cancel so you only have to really multiply the first terms and the last terms.

Let's see:

(a+b)(a-b)

I'm going to use foil:

First:  a(a)=a^2

Outer: a(-b)=-ab

Inner:  b(a)=ab

Last:    b(-b)=-b^2

--------------------------Adding.

a^2-b^2

See -ab+ab canceled so all you had to do was the "first" and "last" of foil.

This would get rid of square roots if a and b had them because they are being squared.

Anyways the conjugate of \sqrt{17}-\sqrt{2} is

\sqrt{17}+\sqrt{2}.

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3 0
3 years ago
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