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Eduardwww [97]
2 years ago
14

The table shows ticket prices for the local minor league baseball team for fanclub members and non-members. For how many tickets

is the cost the same for
club members and non-members?
-

Mathematics
2 answers:
Vladimir79 [104]2 years ago
4 0

Answer with explanation:

Membership fee for club members = $ 30

Ticket Price for for club members =$ 3

Let , number of Club members = x

And,Number of non club members=y

Ticket price for non club members= $ 6

Expressing the above situation in terms of Linear equation

  →30x+3x=6y

   →33x =6 y

Dividing both sides by 3,we get

→11 x=2y

So, when x is multiple of 2=2,4,6,8,...,(For Club Members)

and y is Multiple of 11=11,22,33,44,.....(For Non Club Members)

Ordered pair for Club Members and Non Club Members=(2,11),(4,22),(6,33),......

For these Ordered pairs ,cost is the same for  club members and non-members.

nordsb [41]2 years ago
3 0
The club members would have to buy 2 to have the same amount as one non club members
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Devon wants to write an equation for a line that passes through 2 of the data points he has collected. The points are (8, 5) and
nevsk [136]

Step-by-step explanation:

Points are (8,5) and (-12,-9)

The equation of a line passing through two points is given by :

y-y_1=m(x-x_1)\ \text{m is slope}\\\\y-5=\dfrac{y_2-y_1}{x_2-x_1}(x-8)\\\\y-5=\dfrac{-9-5}{-12-8}\times (x-8)\\\\y-5=\dfrac{7}{10}\times (x-8)\\\\10y-50=7x-56

or

-50+56=7x-10y\\\\7x-10y=6

But he writes the equation 7x – 10y = 3 which is not matching with the one we have calculated. It means that model calculated by Devon is not good.

7 0
2 years ago
The starting salary for a particular job is 1.2 million per annum. The salary increases each year by 75000 to a maximum of 1.5mi
Vlada [557]
<h2>Answer:</h2>

In the 5th year

<h2>Step-by-step explanation:</h2>

For the first year, the salary is 1.2million = 1,200,000

For the second year, the salary is 1.2million + 75000 = 1,200,000 + 75,000 = 1,275,000

.

.

.

For the last year, the salary is 1.5million = 1,500,000

This gives the following sequence...

1,200,000 1,275,000  .   .   . 1,500,000

This follows an arithmetic progression with an increment of 75,000.

<em>Remember that,</em>

The last term, L, of an arithmetic progression is given by;

L = a + (n - 1)d           ---------------(i)

<em>Where;</em>

a = first term of the sequence

n = number of terms in the sequence (which is the number of years)

d = the common difference or increment of the sequence

<em>From the given sequence,</em>

a = 1,200,000                          [which is the first salary]

d = 75,000                               [which is the increment in salary]

L = 1,500,000                          [which is the maximum salary]

<em>Substitute these values into equation (i) as follows;</em>

1,500,000 = 1,200,00 + (n - 1) 75,000

1,500,000 - 1,200,000 = 75,000(n-1)

300,000 = 75,000(n - 1)

\frac{300,000}{75,000} = n - 1

4 = n - 1

n = 5

Therefore, in the 5th year the maximum salary will be reached.

3 0
2 years ago
Xy+(4(20))&gt;x-5y(2+9-7)
prohojiy [21]

I'll solve for y xy+(4(20))>x-5y(2+9-7) 

xy+4(20)>x-5y(2+9-7) 

xy+4(20)>x-5y(4) 

xy+4(20)>x-20y 

xy+80>x-20y 

xy+20y+80>x 

y(x+20)>-80+x 

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3 0
2 years ago
A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

3 0
2 years ago
2 Points
My name is Ann [436]
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3 0
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