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Tpy6a [65]
4 years ago
6

Which statements are true regarding the diagram?

Mathematics
2 answers:
Allushta [10]4 years ago
8 0
Can you show me the diagram
Hatshy [7]4 years ago
3 0

Answer:

A,C,D

Step-by-step explanation:

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I need help factoring this expression
omeli [17]
X^2(x+2)-4(x+2)

(x^2-4) (x+2) = 0

do you have to keep going/?

or thats all they want?
3 0
3 years ago
PLEASE HELP!!!
erastovalidia [21]

Step-by-step explanation:

to write a line equation we need minimum of two points. basicaly a line is written in the form y=mx+c where, m is the slope of the line and c is the intercept made by the line on x-axis (OR) or if two points (x1,y1) and (x2,y2) are given then we can form a line equation as (y-y1)= (y2-y1)*1/(x2-x1) *x-x1 (OR) (y-y2)= (y2-y1)*1/(x2-x1) *x-x2

7 0
3 years ago
If you have 470 apples and the price is 67 apples how much is each apple
EleoNora [17]

Answer:

Each apples cost about $0.14

Step-by-step explanation:

cost / amount = price per piece

67 / 470 = 0.14

So every apple costs about $0.14

8 0
3 years ago
Whole number 6 is equal to ?/3
Ugo [173]

Step-by-step explanation:

That means 6/3 is equivalent to 2 wholes! When the numerator is divisible by the denominator, the fraction is equivalent a whole number. If the numerator is equal to the denominator, you will always have one whole, or 1.

5 0
3 years ago
Simplify the expression. tan(sin^−1 x)
Blizzard [7]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2799412

_______________


Let  \mathsf{\theta=sin^{-1}(x)\qquad\qquad-\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2}.}

(that is the range of the inverse sine function).


So,

\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}


Square both sides:

\mathsf{sin^2\,\theta=x^2\qquad\qquad(but~sin^2\,\theta=1-cos^2\,\theta)}\\\\ \mathsf{1-cos^2\,\theta=x^2}\\\\ \mathsf{1-x^2=cos^2\,\theta}\\\\ \mathsf{cos^2\,\theta=1-x^2}


Since \mathsf{-\,\dfrac{\pi}{2}\ \textless \ \theta\ \textless \ \dfrac{\pi}{2},} then \mathsf{cos\,\theta} is positive. So take the positive square root and you get

\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\quad(ii)}


Then,

\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}


I hope this helps. =)


Tags:  <em>inverse trigonometric function sin tan arcsin trigonometry</em>

3 0
3 years ago
Read 2 more answers
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