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maria [59]
3 years ago
5

Write 75m:125cm in its simplest form​

Mathematics
2 answers:
Ganezh [65]3 years ago
5 0

Answer: should be 3/5

Step-by-step explanation:

Divide both the numerator and denominator by the GCD

75 ÷ 25

125 ÷ 25 if im wrong sorry

Tanya [424]3 years ago
3 0

Answer:

3/5

Step-by-step explanation:

Reduced fraction:

3

5

Therefore, 75/125 simplified to lowest terms is 3/5.

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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
3 years ago
Given: NM || PO and Angle 1 is congruent To angle 3
borishaifa [10]

Answer:

i. LM || NO  (converse alternate interior angle theorem)

ii. <1 ≅ <2    (alternate interior angle theorem)

Step-by-step explanation:

Two or more lines are said to be parallel if they do not meet when extended, even till infinity.

Alternate angles are said to be equal in measure.

Given that;

<1 ≅ < 3,

Since <2 ≅ <3 (alternate interior angle theorem)

Then,

<1 ≅ <2 (transitive property)

Also,

<1 ≅ <2 (alternate interior angle theorem)

Therefore since <1 ≅ <2, thus;

LM || NO (converse alternate interior angle theorem)

7 0
3 years ago
Anne made a recipe for miniature bread loaves that called for StartFraction 7 Over 8 EndFraction of a cup of whole-wheat flour.
Anika [276]

Answer:49/8

Step-by-step explanation:7/8 times 7 is 49/8

6 0
3 years ago
Read 2 more answers
Please help. Could you explan your answer also? thanks
natulia [17]
Answer: Choice B) -4

The lower quartile, also known as Q1 or the first quartile, is the left edge of the box. In this case, that is at -4. We can drop a vertical line from the left edge of the box until we hit -4 on the number line. 

Side Note: 25% of the data values are below Q1, while 75% of the data values are above Q1
4 0
3 years ago
Which is equivalent to square root 10st x square root 15tu
aivan3 [116]

\sqrt{10st}\cdot\sqrt{15tu}=\sqrt{10st\cdot15tu}=\sqrt{150st^{2}u}=\sqrt{25\cdot6\cdot st^{2}u} =   \\\\= \sqrt{5^{2} \cdot t^{2}\cdot6 \cdot su}=5t\sqrt{6su}

4 0
3 years ago
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