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3 years ago

the tickets are 90 dollars in all adults 7 kids 12 kids are 2/3 cost of adults tickets. What is the cost of the adult tickets

Mathematics

1 answer:

svetoff [14.1K]3 years ago

3 0

I hope it helps. .........

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Jan spends part of her year as a member of a gym. She then finds a better deal at another gym, so she cancels her membership wit

valkas [14]

She spent 8 months in the 1st gym and 4 months in the 2nd gym

Step-by-step explanation:

Jan spends part of her year as a member of a gym.

She then finds a better deal at another gym so she cancels her membership with the first gym after x months
She spends the rest of the year, y months, with the second gym
The membership to the first gym costs $75 per month, while the membership for the second gym costs $45 per month
She ended up spending a total of $780 over the course of the year

We need to find how much time she spent at each gym

∵ She spent x months in the 1st gym

∵ She spent y months in the 2nd gym

∵ Her course is a year

- There are 12 months in a year

∴ x + y = 12 ⇒ (1)

∵ The membership to the first gym costs $75 per month

∵ The membership to the second gym costs $45 per month

∵ She ended up spending a total of $780 over the course

∴ 75x + 45y = 780 ⇒ (2)

Now we have a system of equations to solve it

Multiply equation (1) by -45 to eliminate y

∵ -45x - 45y = -540 ⇒ (3)

- Add equations (2) and (3)

∴ 30x = 240

- Divide both sides by 30

∴ x = 8

- Substitute the value of x in equation (1) to find y

∵ 8 + y = 12

- Subtract 8 from both sides

∴ y = 4

She spent 8 months in the 1st gym and 4 months in the 2nd gym

Learn more:

You can learn more about the system of equations in brainly.com/question/2115716

#LearnwithBrainly

4 0

3 years ago

Loretta is trying to build the largest taco in the world she uses 2000 pounds of ground beef one tenth as many pounds of cheese

lakkis [162]

Answer:

2000

Step-by-step explanation:

3 0

2 years ago

The following cube has a volume of 8000 cm3. Find the length of one side of the cube.

Artemon [7]

<span>Vcube = Ledge^3
8000cm^3 = Ledge^3

use the cube root of each side and you have the measure of an edge in cm.

8000^(1/3) = 20
every edge is 20 cm

20^2 = 400
The area of each face is 400 cm^2. </span>

6 0

2 years ago

a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space

agasfer [191]

The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are if $v = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V$

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name $v = a * x ^ 2 +bx +c$ the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently $0 = O + v = (O x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= 0 If O is the neuter, then it ought to restore x², but $0+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.$ This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant

have additive inverse

$Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^ x^ 2 +(vl^ × wl)^ x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)$

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 then $v = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1$

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.

$(1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 3\\1^ × (x^ 2 +x)+2^ × (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )\\(1^ ×2)^ ×(x^ 2 +x)=2^ × (x ^ 2 + x) = 2x + 2\\1^ × (2^ × (x ^ 2 + x) )=1^ × (2x+2)=2x^ 2 +2x( ne2x+2)$

Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently $0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= zero If O is the neuter, then it ought to restore x², but zero $+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.$ This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse

Let $r= v × 2x ^ 2 + v × 1x +v0 , w= w2x ^ 2 + w × 1x +w0 . \\We have thatv+w= (vO + wO) ^ x^ 2 +(vl^ wl)^x+ ( v 2^ w2)w+v= (wO + vO) ^ x^ 2 +(wl^ vl)x+ ( w 2^v2)$

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use $z = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.$ then the associativity rule doesnt hold.

Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

$· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)$