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Sonbull [250]
3 years ago
9

Which Of the following represnt the range of the function y=-1/2(x+10)^2+14?

Mathematics
2 answers:
timofeeve [1]3 years ago
6 0

Answer: the second one

Step-by-step explanation:

cuz i say so

Nonamiya [84]3 years ago
4 0

Answer:

4). But the more correct answer is; y is less than or equal to 14.

Step-by-step explanation:

Graphing this function shows a downward facing parabola with the vertex at (-10,14).  The domain must be less than or equal to 14 because it's all values including and below the vertex since there is negative a value (-1/2).

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2. Solve your equation using PEMDAS
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3. Simplify equation. 
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3 years ago
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

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3 years ago
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ohaa [14]

Answer:

<h2>1) y = 5</h2>

Step-by-step explanation:

x = 4 it's a vertical line.

Perpendicular line to a vertical line is a horizontal line.

A horizontal line has equation y = a.

A horizontal line passes through the point (3, 5) → x = 3 and y = 5.

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VLD [36.1K]

Answer:

I will attach the missing drawing with the answer.

9.b)

Plane JKM

Plane JLM

Plane KLM

Step-by-step explanation:

The drawing for this question is missing. I will attach it with the answer.

9.a) Plane JKL is not an appropriate name for the plane because all of three points lie in the same line.

Through a line pass infinite planes. The plane JKL doesn't define a unique plane. That's why plane JKL isn't an appropriate name for the plane.

9.b) We can name the plane using three points that don't lie in the same line.

Three possible names for the plane are :

Plane JKM

Plane JLM

Plane KLM

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