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3 years ago

41 coins, consisting of nickels and quarters, have a total value of $4.65. How many coins of each kind are there?

Mathematics

2 answers:

slavikrds [6]3 years ago

8 0

Answer:

Step-by-step explanation:

Sergeu [11.5K]3 years ago

3 0

Answer: 28 nickels and 13 quarters

Step-by-step explanation:

Let the number of nickels represented by x and the number of quarters represented by y.

Then according to the question, we have the following equations :

$x+y=41....................(1)\\\\0.05x+0.25y=4.65.................(2)$

Divide 0.05 on both sides of equation (2), we get

$x+5y=93........................(3)$

Subtract equation (1) from equation (3) , we get

$4y=52\\\\\Rightarrow\ y=13$

From (1), x=28

Therefore, there are 28 nickels and 13 quarters.

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3 years ago

Consider the accompanying set of dependent and independent variables. The variable x3 is qualitative with H=High, M=Medium, an

Dominik [7]

Answer:

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Step-by-step explanation:

y 63 53 77 53 38 43 34 20 13

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X3 variable may be represented as thus;

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5 0

3 years ago

HELPPP!!!

spayn [35]

Answer:

A. see below for a graph

B. f(x, y) = f(0, 15) = 90 is the maximum point

Step-by-step explanation:

A. See below for a graph. The vertices are those defined by the second inequality, since it is completely enclosed by the first inequality: (0, 0), (0, 15), (10, 0)

B. For f(x, y) = 4x +6y, we have ...

f(0, 0) = 0

f(0, 15) = 6·15 = 90 . . . . . the maximum point

f(10, 0) = 4·10 = 40

_____

Comment on evaluating the objective function

I find it convenient to draw the line f(x, y) = 0 on the graph and then visually choose the vertex point that will put that line as far as possible from the origin. Here, the objective function is less steep than the feasible region boundary, so vertices toward the top of the graph will maximize the objective function.