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pickupchik [31]
3 years ago
13

What is the measure of the angle shown ?

Mathematics
2 answers:
netineya [11]3 years ago
7 0

Answer:

A. 55°

Step-by-step explanation:

Yuliya22 [10]3 years ago
6 0

Answer:

55 degree

Step-by-step explanation:

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If a given data point is (1,4) and the line of best fit is y = 1.5x + 3.25, what's the residual value?
Svetlanka [38]

Answer:

The residual value is -0.75

Step-by-step explanation:

we know that  

The residual value is the observed value minus the predicted value.

RESIDUAL VALUE=[OBSERVED VALUE-PREDICTED VALUE]

where

Predicted value.--> the predicted value given the current regression equation

Observed value. --> The observed value for the dependent variable.

in this problem

we have the point (1,4)

so

The observed value is 4

<em>Find the predicted value  for x=1 </em>

y =1.5(1)+3.25=4.75

predicted value is 4.75

so

RESIDUAL VALUE=(4-4.75)=-0.75

4 0
3 years ago
The diagonals of a rectangle PQRS intersect at O if angle ROQ=60°
Nitella [24]

Answer:

63783657865781645871661746656395761497567945659871264558716 million answers are possible but ill give u the most accurete answer its f659 degrees

Step-by-step explanation:

6 0
2 years ago
Ned wrote the following statement:
tigry1 [53]

Answer:

You didnt show the statements...

Step-by-step explanation:

7 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
Answer please!! <br> I really need help with thissssssssss! :(
andriy [413]

Answer:

It's the Best answer B

Step-by-step explanation:

Because yes is línea test

You no understand put the B answer friend

6 0
2 years ago
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