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Tanzania [10]
3 years ago
5

PLEASE HELP

Mathematics
2 answers:
algol133 years ago
7 0

Set up a system of equations.


x+y=110

35x + 50y = 5100


Solve for a variable:

y=110-x


Plug this back into the other equation

35x + 50(110-x)=5100

Solve

35x + 5500 - 50x = 5100

-15x = -400

X = 26.66


Plug that into first equation to solve for y:


X+y=110

26.6+y=110

Y=110-26.6

Y=83.4


26.6 small boxes, 83.4 large

Mandarinka [93]3 years ago
5 0

Let x=large boxes and y=small boxes.

x+y=110

60x+35y=5100

____________

You need to cancel out one of the variables to solve for one. Let's cancel out y.

-35(x+y=110)


-35x-35y=-3850

60x+35y=5100

____________

Add the two equations.

25x=1250

x=50

_____________

Plug in 50 for x into one of the equations to solve for y.

50+y=110

y=60

_____________

50 large boxes

60 small boxes

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Answer:

The answer is n=28\frac{11}{15}.

Step-by-step explanation:

Given:

18 2/5 - n = -10 1/3.

Now, to solve the equation:

18\frac{2}{5} - n = -10 \frac{1}{3}

Converting the mixed fractions to improper:

⇒\frac{92}{5} -n=-\frac{31}{3}

Moving the numbers on one side and variable to the other:

⇒\frac{92}{5} +\frac{31}{3}=n

Now, solving the factions:

⇒\frac{92\times 3+31\times 5}{15}=n

⇒\frac{276+155}{15}=n

⇒\frac{431}{15}=n

⇒28\frac{11}{15}=n

Therefore, the answer is n=28\frac{11}{15}.

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ycow [4]

Answer:

A , B, and D

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Answer:

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