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Mekhanik [1.2K]

2 years ago

8

A snack stand sold hot dogs and chips at a recent game. The hot dogs cost $2.50 and the snacks cost $0.75. After the game, they

tallied 175 transactions totaling $262.50.
Represent the linear system in an augmented matrix:
ANSWER IS
[ 1 ] [ 1 ] [ 75 ]
[ 2.5 ] [ 0.75 ] [ 262.5 ]

2 answers:

Paul [167]2 years ago

8 0

Answer:

[1] [1] [175]

[2.5] [0.75] [262.5]

# of hotdogs sold was 75

# of chips sold was 100

Step-by-step explanation:

got it right on edge ya digg

zalisa [80]2 years ago

6 0

Answer:

this is correct on edge

Step-by-step explanation:

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Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

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Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

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A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

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$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

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$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

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$A = 1000000e^{kt}$

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$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

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$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

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3 years ago