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alexdok [17]
4 years ago
10

A large cheese pizza costs $18.Each topping you add on costs $ 1.50. How much would it cost to get a large cheese pizza with c t

oppings added?
Mathematics
2 answers:
mariarad [96]4 years ago
7 0
18 + 1.50c = $ You just make an equation then replace c with the number of toppings
zheka24 [161]4 years ago
6 0
$18*1.50c this is the w=awser
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Fittoniya [83]
It is 546.4583333333333, which would simplify to 546.46 if to the hundredths place or 546.5 to the tens place
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I will mark you brainiest if you can answer this
Klio2033 [76]

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B

Step-by-step explanation:

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PLEASE HELP VERY URGENT ITS DUE TODAY PLEASE HELP Complete the table for the given rule. Rule: y=1/4x + 1
harkovskaia [24]

Answer: Y = 2 ; 3 ; 4

Step-by-step explanation:

Given the rule : y=1/4x + 1

X = 4, 8, 12

When x = 4

y = 1/4(4) + 1

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Therefore ;

X - - - 4 - - - 8 - - - 12

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5 0
3 years ago
What is the volume of a cylinder using pi or 3.14 which the height being 8 and the width being 2​
antoniya [11.8K]

Answer:

The volume of the cylinder is 25.12

Step-by-step explanation:

The width is the diameter in the circle of the base

radius = half of diameter

r = 2/2 = 1

To calculate the volume of a cylinder we have to use the following formula:

v = volume

h = height = 8

π = 3.14

r = radius = 1

v = (π * r²) * h

we replace the unknowns with the values we know

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4 0
3 years ago
Let V denote the set of ordered triples (x, y, z) and define addition in V as in
icang [17]

Answer:

a) No

b) No

c) No

d) No

Step-by-step explanation:

Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:

1. u+v∈V

2. u+v=v+u

3. (u+v)+w=u+(v+w).

4. Exist 0∈V such that u+0=u

5. For each u∈V exist −u∈V such that u+(−u)=0.

6. if c is an escalar and u∈V, then cu∈V

7. c(u+v)=cu+cv

8. (c+d)u=cu+du

9. c(du)=(cd)u

10. 1u=u

let's check each of the properties for the respective operations:

Let u=(u_1,u_2,u_3), v=(v_1,v_2,v_3)

Observe that  

1. u+v∈V

2. u+v=v+u, because the adittion of reals is conmutative

3. (u+v)+w=u+(v+w). because the adittion of reals is associative

4. (u_1,u_2,u_3)+(0,0,0)=(u_1+0,u_2+0,u_3+0)=(u_1,u_2,u_3)

5. (u_1,u_2,u_3)+(-u_1,-u_2,-u_3)=(0,0,0)

then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.

a)

6. c(u_1,u_2,u_3)=(cu_1,u_2,cu_3)\in V

7.

c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),u_2+v_2,c(u_3+v_3))\\=(cu_1+cv_1,u_2+v_2,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv

8.

(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,u_2,(c+d)u_3)=\\=(cu_1+du_1,u_2,cu_3+du_3)\neq (cu_1+du_1,2u_2,cu_3+du_3)=cu+du

Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(ax,y,az)

b)  6. c(u_1,u_2,u_3)=(cu_1,0,cu_3)\in V

7.

c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),0,c(u_3+v_3))\\=(cu_1+cv_1,0,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv

8.

(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,0,(c+d)u_3)=\\=(cu_1+du_1,0,cu_3+du_3)=(cu_1,0,cu_3)+(du_1,0,du_3) =cu+du

9.

c(du)=c(d(u_,u_2,u_3))=c(du_1,0,du_3)=(cdu_1,0,cdu_3)=(cd)u

10

1u=1(u_1,u_2,u3)=(1u_1,0,1u_3)=(u_1,0,u_3)\neq(u_1,u_2,u_3)

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(ax,0,az)

c) Observe that 1u=1(u_1,u_2,u3)=(0,0,0)\neq(u_1,u_2,u_3)

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(0,0,0).

d)  Observe that 1u=1(u_1,u_2,u3)=(2*1u_1,2*1u_2,2*1u_3)=(2u_1,2u_2,2u_3)\neq(u_1,u_2,u_3)=u

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(2ax,2ay,2az).

8 0
3 years ago
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