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AnnyKZ [126]
3 years ago
7

A paper company needs to ship paper to a large printing business. The paper will be shipped in small boxes and large boxes. The

volume of each small box is 10 cubic feet and the volume of each large box is 20 cubic feet. There were twice as many small boxes shipped as large boxes shipped and the total volume of all boxes was 240 cubic feet. Graphically solve a system of equations in order to determine the number of small boxes shipped,
x
,
x, and the number of large boxes shipped,
y
y.
A paper company needs to ship paper to a large printing business. The paper will be shipped in small boxes and large boxes. The volume of each small box is 10 cubic feet and the volume of each large box is 20 cubic feet. There were twice as many small boxes shipped as large boxes shipped and the total volume of all boxes was 240 cubic feet. Graphically solve a system of equations in order to determine the number of small boxes shipped,
x
,
x, and the number of large boxes shipped,
y
y.

Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

  (x, y) = (12, 6)

Step-by-step explanation:

The problem is solved much more easily by considering groups of 2 small and 1 large boxes. Such a group will have a volume of 2·10+20 = 40 ft³, so 240/40 = 6 such groups are needed.

The number of small boxes shipped is 6·2 = 12; the number of large boxes shipped is 6.

_____

To solve this graphically, we must write two equations in x and y:

  x = 2y . . . . . the number of small boxes is twice the number of large boxes

  10x +20y = 240 . . . . the total volume is 240 ft³

The attached graph shows the solution of these equations.

  (x, y) = (12, 6)

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snow_tiger [21]

Answer:

Below, depends if 27 is term number 1 or term number 0.  Answered for both cases.  

Step-by-step explanation:

The most common sequences are arithmetic and geometric, so lets check those first.

Arithmetic first since its the easiest.

to go from 27 to 21 we subtract 6, if we subtract 6 from 21 again we get to 15, which is what we need, so it is indeed arithmetic.

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3 years ago
One morning, John drive 5 hours before stopping to eat. After lunch, he. increased his speed by 10 mph. If he completed a 500-mi
Marrrta [24]
So  he drove 5hrs, before lunch, now, he was going at speed, say "r", after he ate, he sped up by 10mph, so, his rate is whatever "r" is, plus 10, or " r + 10 ", after lunch

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\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before lunch}&d&r&5\\
\textit{after lunch}&500-d&r+10&5
\end{array}
\\\\\\

\begin{cases}
\boxed{d}=5r\\
500-d=(r+10)5\\
----------\\
500-\boxed{5r}=(r+10)5
\end{cases}
\\\\\\
\cfrac{500-5r}{5}=r+10\implies 100-r=r+10

and I'm sure you know what that is
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