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klemol [59]
3 years ago
8

A discount store marks down all of its holiday merchandise by 20% off the regular selling price. Find the discounted price of de

corations that regularly sell for $16 and $23. Write an equation to represent this situation and use it to answer the question.
Mathematics
1 answer:
Anna007 [38]3 years ago
5 0

Answer:

1. $16 = $320

2. $23 = $460

Step-by-step explanation:

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At what point does she lose contact with the snowball and fly off at a tangent? That is
postnew [5]

Answer:

α ≥ 48.2°

Step-by-step explanation:

The complete question is given as follows:

" A skier starts at the top of a very large frictionless snowball, with a very small initial speed, and skis straight  down the side. At what point does she lose contact with the snowball and fly off at a tangent? That is, at the  instant she loses contact with the snowball, what angle α does a radial line from the center of the snowball to  the skier make with the vertical?"

- The figure is also attached.

Solution:

- The skier has a mass (m) and the snowball’s radius (r).

- Choose the center of the snowball to be the zero of gravitational  potential. - We can look at the velocity (v) as a function of the angle (α) and find the specific α at which the skier lifts off and  departs from the snowball.

- If we ignore snow-­ski friction along with air resistance, then the one work producing force in this problem, gravity,  is conservative. Therefore the skier’s total mechanical energy at any angle α is the same as her total mechanical  energy at the top of the snowball.

- Hence, From conservation of energy we have:

                       KE (α) + PE(α) = KE(α = 0) + PE(α = 0)

                       0.2*m*v(α)^2 + m*g*r*cos(α) = 0.5*m*[ v(α = 0)]^2 + m*g*r

                       0.2*m*v(α)^2 + m*g*r*cos(α) ≈ m*g*r

                        m*v(α)^2 / r = 2*m*g( 1 - cos(α) )

- The centripetal force (due to gravity) will be mgcosα, so the skier will remain on the snowball as long as gravity  can hold her to that path, i.e. as long as:

                         m*g*cos(α) ≥ 2*m*g( 1 - cos(α) )

- Any radial gravitational force beyond what is necessary for the circular motion will be balanced by the normal  force—or else the skier will sink into the snowball.

- The expression for α_lift becomes:

                            3*cos(α) ≥ 2

                            α ≥ arc cos ( 2/3) ≥ 48.2°

4 0
2 years ago
Solve the equation. |4r - 16| = 12<br> i need help asap
soldi70 [24.7K]
I believe the answer is 7
3 0
3 years ago
An envelope contain 40 shopping vounchers of which 25 voucher each have value of $50 and 15 vouchers each have a value of $100.
Ghella [55]

Answer:

yes is correct ◆◆◆◆◆◆◆◆◆◆♡♡♡♡♡

7 0
2 years ago
Read 2 more answers
What is the vertex of the graph of the function y=-2x^2+16x-15
andrey2020 [161]
<u>X - Vertex</u>
Vertex = -\frac{b}{2a} \\Vertex = -\frac{16}{2(-2)} \\Vertex = -\frac{16}{-4} \\Vertex = 4

<u>Y - Vertex</u>
y = -2x^{2} + 16x - 15 \\y = -2(4)^{2} + 16(4) - 15 \\y = -2(16) + 64 - 15 \\y = -32 + 64 - 15 \\y = 32 - 15 \\y = 17

Vertex:(4, 17)
8 0
3 years ago
Carl is wondering if the train he is riding home from school will leave early, on time, or late. The probabilities are as follow
Slav-nsk [51]

Answer:

Probability of each situation is 1/3 .

Step-by-step explanation:

There are 3 possibilities

1- Train will arrive early

2- Train will arrive on time

3- Train will arrive late

Formula for Probability  of event a is = n(a)/Sum of events

In this case sum of events = 3

So

Probability of early arrival = 1/3

Probability of on time arrival = 1/3

Probability of late arrival = 1/3

5 0
3 years ago
Read 2 more answers
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