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.

$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.

$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 :-

                           (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

  (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

Put value of c from (2)

Subtract 14 from both sides , we get

Divide both sides by 2 , we get

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

At x= 5

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

160

step-by-step explanation:

idk

i hope i

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.