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UNO [17]
3 years ago
12

A store sells Brazilian coffee for $10 per lb. and Columbian coffee for $14 per lb. If the store decides to make a 150-lb. blend

of the two and sell it for $11 per lb., how much of each type of coffee should be used?
Mathematics
1 answer:
Artemon [7]3 years ago
6 0

Answer:

The store should use 112.5 pounds of Brazilian coffee and 37.5 pounds of Colombian cofee.

Step-by-step explanation:

Let "b" be the amount of Brazilian coffee, in pounds, required for the blend and "c" the amount of Colombian coffee required, in pounds.

Since there are two unknown variables a two-equation system is needed to solve the problem, we can set up one equation for weight and another for price as follows:

b+c=150\\10*b+14*c=11*150

Solve for "c" by multiplying the first equation by -10 and adding it to the second one:

b+c=150\\10b+14c -10b-10c=11*150 -(10*150)\\4c=150\\c = 37.5

Now, solve for b by replacing the value obtained into the first equation

b+c=150\\b= 150 - 37.5\\b=112.5

The store should use 112.5 pounds of Brazilian coffee and 37.5 pounds of Colombian cofee.

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Answer:

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Integers are just numbers that doesn't have a decimal point.

1, 2, 3, 4, and 5 are integers.

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4:0.5

8:1

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16:2

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I need Help! Look at equation A,B,andC
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Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th
zalisa [80]
We have the following equations:

(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

y \leq -0.5x+5 \\ y \leq -1.25x+8

Region B.

Let's take a point that is in this region, that is:

P(0,6)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5  \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\  \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8

So the inequalities are:

(1) \ y  \geq  -0.5x+5 \\  (2) \ y  \leq  -1.25x+8

Region C.

A point in this region is:

P(0,10)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \  \ 8

So the inequalities are:

(1) \ y  \geq  -0.5x+5 \\ (2) \ y  \geq  -1.25x+8

Region D.

A point in this region is:

P(8,0)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \  \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2

So the inequalities are:

(1) \ y  \leq  -0.5x+5 \\ (2) \ y  \geq  -1.25x+8

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\  \\ Solving \ for \ y: \\ y=-0.5(4)+5=3

Accordingly, the point is:

\boxed{p(4,3)}

Point q:

This point is the x-intercept of the line:

y=-0.5x+5

So let:

y=0

Then

x=\frac{5}{0.5}=10

Therefore, the point is:

\boxed{q(10,0)}

Part b) 

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

O(0,0)

Substituting in the inequalities:

y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}

It is true.

Region B.

Let's take a point that is in this region, that is:

P(0,7)

Substituting in the inequalities:

y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y  \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}

It is true

Region C.

Let's take a point that is in this region, that is:

P(0,11)

Substituting in the inequalities:

y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}

It is true

Region D.

Let's take a point that is in this region, that is:

P(9,0)

Substituting in the inequalities:

y  \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}

It is true

7 0
3 years ago
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