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Rufina [12.5K]
3 years ago
13

What best describes how to evaluate a variable expression?

Mathematics
1 answer:
Kaylis [27]3 years ago
7 0
<span> To replace each letter with its value, and then finish it by doing the order of the operation/ solve it.</span>
You might be interested in
Distance = (9-3)2+(-2-6)2
Hoochie [10]

Answer:

-5

Step-by-step explanation:

Its negitive 5 because if you add all that up then it all equals -5. Sorry im not really good at evplaining things but the distance is -5.

6 0
2 years ago
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
3 years ago
Pls help ill give brainliest ​
Maurinko [17]

Answer:

4

Step-by-step explanation:

-5g -6

Let g= -2

-5 (-2) -6

Multiply

10 -6

4

7 0
2 years ago
Read 2 more answers
Write an expression for “the quotient for y and 8”
jonny [76]

Answer:


Step-by-step explanation:

I think your answer is y/8. It's a little unusually worded. If this is not correct, please leave a note.

The way I have answered it, I would word it as "What is the quotient of y divided by 8."


7 0
3 years ago
If 2% of a number is 8 what is that number---please explain
aniked [119]
~Hello There!~

Multiplier = 0.02
8/0.02 = 400
The number is 400.

Hope This Helps You!
Good Luck :)
Have A Great Day ^_^

- Hannah ❤
5 0
3 years ago
Read 2 more answers
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