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Tamiku [17]
2 years ago
5

Maya had $27. She spent all the money on buying 3 burgers for $x each and 2 sandwiches for $y each. If Maya would have bought 2

burgers and 1 sandwich, she would have been left with $11. A student concluded that the price of each burger is $5 and the price of each sandwich is $6. Which statement best justifies whether the student's conclusion is correct or incorrect?
The student's conclusion is correct because the solution to the system of equations 3x + 2y = 11 and 2x + y = 16 is (5, 6).
b. The student's conclusion is incorrect because the solution to the system of equations 3x - 2y = 11 and 2x - y = 16 is (5, 6).
c. The student's conclusion is correct because the solution to the system of equations 3x + 2y = 27 and 2x + y = 16 is (5, 6).
d. The student's conclusion is incorrect because the solution to the system of equations 2x + 3y = 27 and x + 2y = 16 is (5, 6).
Mathematics
2 answers:
yKpoI14uk [10]2 years ago
7 0
C is the correct answer. Solve for Y on the second equation, plug it in the other equation, and solve for x.
defon2 years ago
6 0

Answer:

Option C

The student's conclusion is correct because the solution to the system of equations 3x+2y=27  and 2x+y=16 is (5,6)

Step-by-step explanation:

Let

x------> the cost of each burger

y-----> the cost of each sandwich

we know that

3x+2y=27 -----> equation A

2x+y=(27-11)

2x+y=16 ----> equation B

Solve the system of equations

Multiply the equation B by -2

-2(2x+y)=-2*16 ----->-4x-2y=-32 ------> equation C

Adds equation A and equation C

3x+2y=27 \\-4x-2y=-32\\ ----------\\ 3x-4x=27-32\\ -x=-5\\x=5

Substitute the value of x in the equation B

2(5)+y=16

10+y=16

y=6

The solution of the system of equations is the point (5,6)

so

The price of each burger is \$5

The price of each sandwich is \$6

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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)}


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. =)

3 0
3 years ago
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