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3 years ago

Can you help me is monomials

Mathematics

1 answer:

Novay_Z [31]3 years ago

5 0

Section 2.4

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8x^2y : 2x : 2y = 2x

This is because 8x^2y divided over 2x leads to 4xy. Dividing 4xy over 2y leads to 2x.

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15x^6y^3 : 5y^2x^3 = 3x^3y

Note how 15 divided by 5 is 3

Also, x^6 divided by x^3 is x^3. You subtract the exponents.

Similarly, y^3 divided by y^2 is y^1 = y.

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16y^5x^3z : -4zyx = -4y^4x^2

Divide the coefficients to get 16/(-4) = -4

The y terms divide to (y^5)/(y) = y^4. Subtract the exponents.

The x and z terms are handled the same way.

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-8x^2 : 2x = -4x

We follow the same steps as before. This time -8/2 = -4 is the new coefficient. The x terms divide to (x^2)/x = x where x is nonzero.

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-6xz : -2zx = 3

The coefficients divide to (-6)/(-2) = 3

The x and z terms cancel out when we divide them.

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3y^2x : xy^2 = 3

The y^2x expression is the same as xy^2, so when we divide these terms they cancel.

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The distance would be 408 feet

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3 years ago

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Answer:

The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.

Step-by-step explanation:

Let us take the point of projection of the ball as origin of the coordinate system, the upward direction as positive and down direction as negative.

Initial velocity u with which the ball is projected upwards = + 120 ft/s

Uniform acceleration a acting on the ball is to acceleration due to gravity = - 32 ft/s²

The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.

Using the formula:

v² - u² = 2 a h,

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u = initial velocity of the ball = +120 ft/s

v = final velocity of the ball at the highest point = 0 ft/s

a = uniform acceleration acting on the ball = -32 ft/s²

h = height attained

Substituting the values we get;

0² - 120² = 2 × (- 32) h

=> h = 120²/2 × 32 = 225 feet

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3 years ago

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Answer:

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5 0

2 years ago

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Can you help me is monomials​

Can you help me is monomials