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ANEK [815]
3 years ago
7

Cans of black beans usually cost $1.34 each. This week they are on sale for 5 cans for $6.00. You bought one can last week and o

ne can this week. How much more or less did you pay this week?
Mathematics
1 answer:
mixer [17]3 years ago
7 0

Answer:

<u>$0.14</u> less to pay this week.

Step-by-step explanation:

Given:

Cans of black beans usually cost $1.34 each.

This week they are on sale for 5 cans for $6.00.

You bought one can last week and one can this week.

Now, to find the amount more or less to pay this week.

So, we use unitary method to get the price of can this week:

If, 5 cans cost this week = $6.00.

Then, 1 can cost this week = \frac{6}{5}=\$1.20.

<u><em>Thus, the cans of black beans cost $1.20 this week.</em></u>

<u><em>And, last week the cans cost $1.34 each.</em></u>

<u><em>So, the cost of can this week is less than last week.</em></u>

Now, we get the difference in the cost of can of this week than last week by subtracting the cost of can last week from cost of can last week:

\$1.34-\$1.20\\\\=\$0.14.

Therefore, $0.14 less to pay this week.

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Step-by-step explanation:

It is given that a cab ride cost $2 which is a fixed amount and charges additional $3 per mile. So you have to make an equation of c in terms of d :

Let c be the cost,

Let d be the distance,

c = 2 + 3d

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2 years ago
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Answer = 5.656854249

Step by step:

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Julli [10]

Answer:

Step-by-step explanation:

1= 2/3

2=5/6

3=21/40

4= 2 11/20

5= 1 1/3

6=5/6

7=1 1/2

8=9/10

9= 5 1/6

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2 years ago
Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and
zloy xaker [14]

Answer:

(0,0)   (4000,0) and (500,79)

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW

\frac{dW}{dt} = -0.02W + 0.00004RW

To solve this, we first equate \frac{dR}{dt} and \frac{dW}{dt} to 0.

So, we have:

0.09R(1 - 0.00025R) - 0.001RW = 0

-0.02W + 0.00004RW = 0

Factor out R in 0.09R(1 - 0.00025R) - 0.001RW = 0

R(0.09(1 - 0.00025R) - 0.001W) = 0

Split

R = 0   or 0.09(1 - 0.00025R) - 0.001W = 0

R = 0   or  0.09 - 2.25 * 10^{-5}R - 0.001W = 0

Factor out W in -0.02W + 0.00004RW = 0

W(-0.02 + 0.00004R) = 0

Split

W = 0 or -0.02 + 0.00004R = 0

Solve for R

-0.02 + 0.00004R = 0

0.00004R = 0.02

Make R the subject

R = \frac{0.02}{0.00004}

R = 500

When R = 500, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0

0.09 -0.01125 - 0.001W = 0

0.07875 - 0.001W = 0

Collect like terms

- 0.001W = -0.07875

Solve for W

W = \frac{-0.07875}{ - 0.001}

W = 78.75

W \approx 79

(R,W) \to (500,79)

When W = 0, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0

0.09 - 2.25 * 10^{-5}R = 0

Collect like terms

- 2.25 * 10^{-5}R = -0.09

Solve for R

R = \frac{-0.09}{- 2.25 * 10^{-5}}

R = 4000

So, we have:

(R,W) \to (4000,0)

When R =0, we have:

-0.02W + 0.00004RW = 0

-0.02W + 0.00004W*0 = 0

-0.02W + 0 = 0

-0.02W = 0

W=0

So, we have:

(R,W) \to (0,0)

Hence, the points of equilibrium are:

(0,0)   (4000,0) and (500,79)

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Answer:

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Step-by-step explanation:

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