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abruzzese [7]
3 years ago
14

What 0.003 a tenth of i need a real answer

Mathematics
2 answers:
Andrej [43]3 years ago
5 0
3000th is the answer.
If I get it wrong, I apologize. I didn't exactly understand your grammar..
OverLord2011 [107]3 years ago
3 0
I think 0.003 would be a tenth of 0.03 because you would multiply it by ten, so then when you divide it by ten, you get 0.003
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Use trigonometric identities to solve each equation within the given domain.
katrin [286]

Recall that the tangent function is defined by

tan(<em>x</em>) = sin(<em>x</em>)/cos(<em>x</em>)

Also recall the double angle identity for sine,

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Then the equation is the same as

3 sin(<em>x</em>)/cos(<em>x</em>) = 4 sin(<em>x</em>) cos(<em>x</em>)

Move everything to one side to prepare to factorize:

3 sin(<em>x</em>)/cos(<em>x</em>) - 4 sin(<em>x</em>) cos(<em>x</em>) = 0

sin(<em>x</em>)/cos(<em>x</em>) (3 - 4 cos²(<em>x</em>)) = 0

As long as cos(<em>x</em>) ≠ 0, we can omit the term in the denominator, so we're left with

sin(<em>x</em>) (3 - 4 cos²(<em>x</em>)) = 0

and so

sin(<em>x</em>) = 0   <u>or</u>   3 - 4 cos²(<em>x</em>) = 0

sin(<em>x</em>) = 0   <u>or</u>   cos²(<em>x</em>) = 3/4

sin(<em>x</em>) = 0   <u>or</u>   cos(<em>x</em>) = ±√3/2

On the interval [0, 2<em>π</em>),

• sin(<em>x</em>) = 0 for <em>x</em> = 0 and <em>x</em> = <em>π</em>

• cos(<em>x</em>) = √3/2 for <em>x</em> = <em>π</em>/6 and <em>x</em> = 11<em>π</em>/6

• cos(<em>x</em>) = -√3/2 for <em>x</em> = 5<em>π</em>/6 and <em>x</em> = 7<em>π</em>/6

(None of these <em>x</em> make cos(<em>x</em>) = 0, so we don't have to omit any extraneous solutions.)

6 0
3 years ago
What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC?
natulia [17]

Answer:

<em>The endpoint coordinates for the mid-segment are:  (-2,-1) and (1,0)</em>

Step-by-step explanation:

According to the given diagram, the coordinates of the vertices of \triangle BCD are:   B(-3,1), C(3,3) and D(-1,-3)

Now, the endpoints of the mid-segment of \triangle BCD which is parallel to BC will be the mid-points of sides BD and CD.

<u>Formula for the coordinate of mid-point</u> :   (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}), where (x_{1}, y_{1}) and (x_{2}, y_{2}) are two endpoints.

So, the mid-point of BD will be:  (\frac{-3-1}{2},\frac{1-3}{2})=(\frac{-4}{2},\frac{-2}{2})=(-2,-1)

and the mid-point of CD will be:  (\frac{3-1}{2},\frac{3-3}{2})=(\frac{2}{2},\frac{0}{2})=(1,0)

Thus, the endpoint coordinates for the mid-segment of \triangle BCD that is parallel to BC are  (-2,-1) and (1,0)

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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