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Nutka1998 [239]
2 years ago
10

Cheddar Cheese

Mathematics
1 answer:
disa [49]2 years ago
4 0

Complete question :

Cheddar Cheese

$3/lb

Swiss Cheese

$5/lb

Keisha is catering a luncheon. She has $30 to spend on a mixture of Cheddar cheese and Swiss cheese. How many pounds of cheese can Keisha get if she buys only Cheddar cheese? Only Swiss cheese? A mixture of both cheeses?

What linear equation in standard form can she use to model the situation?

Answer:

10 lbs of cheddar cheese

6 lbs of Swiss cheese

$3a + $5b = $30

Step-by-step explanation:

Given that :

Cheddar cheese = $3/lb

Swiss cheese = $5/lb

Total amount budgeted for cheese = $30

How many pounds of cheese can Keisha get if she buys only Cheddar cheese?

Pounds of cheedar cheese obtainable with $30

Total budget / cost per pound of cheddar cheese

$30 / 3 = 10 pounds of cheedar cheese

Only Swiss cheese?

Pounds of cheedar cheese obtainable with $30

Total budget / cost per pound of Swiss cheese

$30 / 5 = 6 pounds of Swiss cheese

A mixture of both cheeses?

What linear equation in standard form can she use to model the situation?

Let amount of cheddar cheese she can get = a

Let amount of Swiss cheese she can get = b

Hence,

(Cost per pound of cheddar cheese * number of pounds of cheddar) + (Cost per pound of Swiss cheese * number of pounds of Swiss cheese) = total budgeted amount

(3 * a) + (5 * b) = $30

$3a + $5b = $30

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Answer:

<em>Hence the daughter's present age is 15 years</em>

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<em></em>

Step-by-step explanation:

Let the present age of daughter be x

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Daughter's age = x - 5

Fathers age = y - 5

If the present age of father is thrice as old as the age of daughter 5 years ago, then;

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y = 3x - 10 .... 1

In 5 years time;

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Fathers age = y + 5

If the age of father will be twice the age of his daughter in 5 years time then;

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Equate 1 and 2;

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3x - 2x = 5 + 10

x = 15

Since y = 2x + 5

y = 2(15) + 5

y = 35

<em>Hence the daughter's present age is 15 years</em>

<em>The fathers present age is 35 years</em>

<em></em>

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azamat
The total distance was 35 miles. Why? Well, see below:

The student ran 3 miles each day for 5 days. That can be representative of:

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Day 3: 3 miles
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Adding these values together, we will find that the student ran a total of 15 miles in the first five days.

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Answer:

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