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Lunna [17]
2 years ago
8

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

1) The student's conclusion is correct because the solution to the system of equations 3x + 2y = 11 and 2x + y = 16 is (5, 6).
2) The student's conclusion is incorrect because the solution to the system of equations 3x − 2y = 11 and 2x − y = 16 is (5, 6).
3) The student's conclusion is correct because the solution to the system of equations 3x + 2y = 27 and 2x + y = 16 is (5, 6).
4) 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
1 answer:
Ber [7]2 years ago
3 0
Let
x-------> the cost of the burger
y-------> the cost of the <span>sandwiches

we know that
3x+2y=27-----> equation 1
2x+y=27-11-----> 2x+y=16-----> multiply by -2-----> -4x-2y=-32--> equation 2

adds equation 1 and equation 2
</span> 3x+2y=27
-4x-2y=-32
<span>----------------
-x=27-32-----> x=5
3*5+2y=27----> 2y=27-15-----> y=6

therefore

the answer is the option
</span><span>3) The student's conclusion is correct because the solution to the system of equations 3x + 2y = 27 and 2x + y = 16 is (5, 6).</span>
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Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v
Bumek [7]

Complete Question

The  complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

is not dimensionally consistent

E

is not dimensionally consistent

F

is dimensionally consistent

G

is dimensionally consistent

H

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

   The equation are

                        A) \   \  a^3  =  \frac{x^2 v}{t^5}

                       

                       B) \   \  x  =  t

 

                       C \ \ \ v  =  \frac{x^2}{at^3}

 

                      D \ \ \ xa^2 = \frac{x^2v}{t^4}

                      E \ \ \ x  = vt+ \frac{vt^2}{2}

                     F \ \ \  x = 3vt

 

                    G \ \ \  v =  5at

 

                    H \ \ \  a  =  \frac{v}{t} + \frac{xv^2}{2}

Generally in dimension

     x - length is represented as  L

     t -  time is represented as T

     m = mass is represented as M

Considering A

           a^3  =  (\frac{L}{T^2} )^3 =  L^3\cdot T^{-6}

and    \frac{x^2v}{t^5 } =  \frac{L^2 L T^{-1}}{T^5}  =  L^3 \cdot T^{-6}

Hence

           a^3  =  \frac{x^2 v}{t^5} is dimensionally consistent

Considering B

            x =  L

and      

            t = T

Hence

      x  =  t  is not dimensionally consistent

Considering C

     v  =  LT^{-1}

and  

    \frac{x^2 }{at^3} =  \frac{L^2}{LT^{-2} T^{3}}  =  LT^{-1}

Hence

   v  =  \frac{x^2}{at^3}  is dimensionally consistent

Considering D

    xa^2  = L(LT^{-2})^2 =  L^3T^{-4}

and

     \frac{x^2v}{t^4}  = \frac{L^2(LT^{-1})}{ T^5} =  L^3 T^{-5}

Hence

    xa^2 = \frac{x^2v}{t^4}  is not dimensionally consistent

Considering E

   x =  L

;

   vt  =  LT^{-1} T =  L

and  

    \frac{vt^2}{2}  =  LT^{-1}T^{2} =  LT

Hence

   E \ \ \ x  = vt+ \frac{vt^2}{2}   is not dimensionally consistent

Considering F

     x =  L

and

    3vt = LT^{-1}T =  L      Note in dimensional analysis numbers are

                                                       not considered

  Hence

       F \ \ \  x = 3vt  is dimensionally consistent

Considering G

    v  =  LT^{-1}

and

    at =  LT^{-2}T =  LT^{-1}

Hence

      G \ \ \  v =  5at   is dimensionally consistent

Considering H

     a =  LT^{-2}

,

       \frac{v}{t}  =  \frac{LT^{-1}}{T}  =  LT^{-2}

and

    \frac{xv^2}{2} =  L(LT^{-1})^2 =  L^3T^{-2}

Hence

    H \ \ \  a  =  \frac{v}{t} + \frac{xv^2}{2}  is not dimensionally consistent

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Answer:

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Step-by-step explanation:

L(w) = length of beard as a function of time in weeks

L(w) = 8 mm + (2 mm/wk)(wk)

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We will use \frac{28}{36} ratio to determine how much can Fatima afford as a loan for her house.

Let gross income of Fatima = $ x

Amount earned by Fatima monthly= $ 5,375

36 % of 5,375=

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Maximum amount that can be paid as a loan monthly = $ 1935

Total money that she can pay yearly for house loan if Fatima allowable money for paying loans is $ 1935=

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Money saved as a down payment = $ 15,000

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She is not allowed for loan for house if the cost of house is any of the four options.

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