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Naddika [18.5K]
2 years ago
4

The cost in dollars, y, of a large pizza with x toppings from Pat’s Pizzeria can be modeled by a linear function. A large pizza

with no toppings costs $14.00. A large pizza with 2 toppings costs $17.50.
What is the cost of a pizza with 5 toppings? Round to the nearest penny.
Mathematics
2 answers:
Andrei [34K]2 years ago
5 0

$22.75

Start by finding the cost of each topping. Subtract $17.50 minus $14.00 to find that adding 2 toppings costs $3.50. Now, divide $3.50 by 2 to find that the cost for adding only one topping is $1.75.

Then, multiply $1.75 by 5 to find how much it costs to add 5 toppings. You get $8.75.

Finally, add the cost of 5 toppings to the cost of a large pizza with no toppings. $14.00 plus $8.75 equals $22.75, so a large pizza with 5 toppings costs $22.75.

Naya [18.7K]2 years ago
3 0

Answer: $22.75

Step-by-step explanation:

Given : The cost in dollars, y, of a large pizza with x toppings from Pat’s Pizzeria can be modeled by a linear function.

A linear function is given by :-

y=mx+c                            (1)

, where m is slope (rate of change of y w.r.t x) and c is the y-intercept.

A large pizza with no toppings costs $14.00.

i.e. for x=0 , y= 14

Put theses values in (1) , we get

14=m(0)+c\\\Rightarrow\ c=14   (2)

A large pizza with 2 toppings costs $17.50.

i.e. for x=2 , y= 17.50

Put theses values in (1) , we get

17.50=m(2)+c      

Put value of c from (2)

17.50=2m+14\\\\      

Subtract 14 from both sides , we get

3.50=2m

Divide both sides by 2 , we get

1.75=m

Put m= 1.75 and c= 14 in (1) , the linear function representing cost of large pizza becomes y=1.75x+14

At x= 5

y=1.75(5)+14=8.75+14=22.75

Thus , the cost of a pizza with 5 toppings= $22.75

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Answer:

5.024 × 10^8 tons /day

Step-by-step explanation:

This question has to deal with conversion

We are told in the question that:

On average, water flows over a particular water fall at a rate of 2.05 x 10^5 cubic feet per second.

Water flow rate = 2.05 × 10^5 ft³/s

From the question,

One cubic foot of water weighs 62.4 lb.

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Hence,

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2.05 × 10^5 ft³ =

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From the question,

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12792000 Ib =

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E

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