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IrinaK [193]
3 years ago
14

What is the solution to this system of equations?

Mathematics
1 answer:
amm18123 years ago
3 0

Answer:

Step-by-step explanation:

(i) x+2y=4

(ii) x=4-2y

(iii) 2x-2y=5  

substituting (ii) in (iii)

2x-2y=5

2(4-2y)-2y=5

8-4y-2y=5

-6y=5-8

-6y=-3

(iv) y=\frac{1}{2}              

substituting (iv) in (ii)

x=4-2y

x=4-2×\frac{1}{2}

x=4-1

x=3

∴ B.(3,\frac{1}{2})

HOPE IT HELPS YOU!!!!

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To input your angle measures in degrees, use the functions sind, cosd and tand, instead of sin, cos and tan. Zach stands at the
kramer

Answer:

  about 2949 feet

Step-by-step explanation:

The geometry of the situation can be modeled by a right triangle. The height of the cliff can be taken to be the side opposite the given angle, and the distance to the coyote will be the side adjacent to the given angle. The relation between these values is the trig function ...

  Tan = Opposite/Adjacent

__

<h3>setup</h3>

Filling in the known values, we have ...

  tan(6°) = (310 ft)/(distance to coyote)

<h3>solution</h3>

Multiplying by (distance to coyote)/tan(6°) gives ...

  distance to coyote = (310 ft)/tan(6°) ≈ 310/0.105104 ft

  distance to coyote ≈ 2949.453 ft

The coyote is about 2949 feet from the base of the cliff.

6 0
2 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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Answer:

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Step-by-step explanation:

Step 1: Flip the equation.

-4/3s <-4

Step 2: Multiply both sides by 3/(-4).

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dlinn [17]

Answer:

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Step-by-step explanation:

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open the bracket

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Check/ verify

A=1/2(a+b)h

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32=1/2(16)4

divide 16 by 2

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Step-by-step explanation: found it on quizlet

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