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givi [52]
3 years ago
13

PLEASE HELP I HAVE 5 MIN TO TURN THIS IN!!!!

Mathematics
2 answers:
sineoko [7]3 years ago
7 0

Answer:

  • False
  • the equation required to fence both garden do not match the perimeter given on statement 2.
  • 41x - 37 ≠ 41x + 35

Step-by-step explanation:

<u>garden A</u>

Width (W) = 2x - 3

Length (L) = 5(W)

                = 5 (2x - 3)

                = 10x - 15

Perimeter A = 2L + 2W  (perimeter of a rectangular garden)

                    = 2(10x - 15) + 2(2x - 3)

                    = 20x - 30 + 4x - 6

                    = 24x - 36

<u>garden B</u>

side 1 = 5x + 5

side 2 = 4x + 2

side 3 = 8x - 8

Perimeter B = sides (1 + 2 + 3)  

                    = 5x+5 + 4x+2 + 8x-8

                    = 5x + 4x + 8x + 5 + 2 - 8

                    = 17x - 1

Garden A + garden B  = statement 2

24x - 36  +  17x - 1

24x + 17x - 36 - 1

 41x - 37

therefore, the statement 2 is <u>false.</u> as the equation required to fence both garden do not match the perimeter given on statement 2.

<h3>41x - 37 ≠ 41x + 35</h3>

Lady_Fox [76]3 years ago
6 0

Answer:

  • False

Step-by-step explanation:

<u>Perimeter of garden A  is </u>

  • P = 2(w + l) = 2( 2x - 3 + 5(2x - 3)) = 24x - 36

<u>Perimeter of garden B  is </u>

  • 5x + 5 + 4x + 2 + 8x - 8 = 17x - 1

<u>Sum of perimeters:</u>

  • 24x - 36 + 17x - 1 = 41x - 37

The statement 2 is FALSE as number we got is different from the one in the statement

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——————————

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)}


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\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. =)

3 0
3 years ago
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