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emmasim [6.3K]
3 years ago
5

PLEASE HELP! Write an algebraic expression for 9 groups of w

Mathematics
2 answers:
Marysya12 [62]3 years ago
8 0

It is 9w hope this helps


stiks02 [169]3 years ago
7 0
The algebraic expression is 9w.

Hope this helps!
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What is the greatest common factor of 60 and 75? A.3 B.6 C.15 D.25
bekas [8.4K]
The answer is C, 15.
7 0
3 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
What is the area of shaded section of the circle?​
bazaltina [42]

Answer:

≈ 565.5 units²

Step-by-step explanation:

The area (A) of the shaded sector = area of circle × fraction of circle

A = πr² × \frac{162}{360} ← r is the radius

   = π × 20² × \frac{162}{360}

   = 400π × 0.45 ≈ 565.5 units²

4 0
3 years ago
Can someone pls help me with this question ​
jeyben [28]

Answer:

I'll find X for you; the next step is by you.

Step-by-step explanation:

we have two similar pentagons

I will do it in 2 ways.

1.

=> 48/4x = (x+2)/x

=> 12/x = (x+2)/x

Because of pentagons so x > 0

=> 12 = x + 2 => x = 10

2.

k is unknown number

=> 4x = k.x and 48 = k.(x + 2) (x > 0)

=> k = 4 and 48 = 4.(x + 2)

=> 12 = x + 2 => x = 10

7 0
2 years ago
A sum amount <br>to rupees 34476 in two and a half years at 4% CI. the sum is ​
Kruka [31]

Given:

Amount = Rs. 34476

Rate of compound interest = 4%

Time = 2\dfrac{1}{2}=2.5 years

To find:

The principal value.

Solution:

Formula for amount is

A=P\left(1+\dfrac{r}{100}\right)^t

Where, P is principal value, r is rate of interest and t is time in years.

Putting the given values, we  get

34476=P\left(1+\dfrac{4}{100}\right)^{2.5}

34476=P\left(1+0.04\right)^{2.5}

34476=P\left(1.04\right)^{2.5}

\dfrac{34476}{\left(1.04\right)^{2.5}}=P

Now,

P=31256.0090

P\approx 31256

Therefore, the value of sum or principal value is Rs.31256.

6 0
2 years ago
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