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2 years ago

Helloooo can someone help meeeeeeeee

Mathematics

1 answer:

Liono4ka [1.6K]2 years ago

8 0

Answer:

Option B, $(48x^{8}y^{5})$

Step-by-step explanation:

<u>Step 1: Simplify the exponent</u>

$(12x^{2}y^{3})(2x^{3}y)^{2}$

$(12x^{2}y^{3})(2^{2}x^{3*2}y^{1*2})$

$(12x^{2}y^{3})(4x^{6}y^{2})$

<u>Step 2: Simply the expression</u>

$(12x^{2}y^{3}*4x^{6}y^{2})$

$(48x^{2+6}y^{3+2})$

$(48x^{8}y^{5})$

Answer: Option B, $(48x^{8}y^{5})$

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3 years ago

An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or abov

babunello [35]

Answer:

D) 14 seconds

Step-by-step explanation:

First we will plug 500 in for y:

500 = -4.9t² + 120t

We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:

500-500 = -4.9t² + 120t - 500

0 = -4.9t²+120t-500

Our values for a, b and c are:

a = -4.9; b = 120; c = -500

We will use the quadratic formula to solve this. This will give us the two times that the object is at exactly 500 meters. The difference between these two times will tell us when the object is at or above 500 meters.

The quadratic formula is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Using our values for a, b and c,

$x=\frac{-120\pm \sqrt{120^2-4(-4.9)(-500)}}{2(-4.9)}\\\\=\frac{-120\pm \sqrt{14400-9800}}{-9.8}\\\\=\frac{-120\pm \sqrt{4600}}{-9.8}\\\\=\frac{-120\pm 67.8233}{-9.8}\\\\=\frac{-120+67.8233}{-9.8}\text{ or }\frac{-120-67.8233}{-9.8}\\\\=\frac{-57.1767}{-9.8}\text{ or }\frac{-187.8233}{-9.8}\\\\=5.3242\text{ or }19.1656$

The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. This means the amount of time it is at or above 500 meters is

19-5 = 14 seconds.

6 0

3 years ago

What is the answer please im so anyyoned its inquatiltys and greater than less than please help

Basile [38]

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3 years ago

I am so lost, please help

skelet666 [1.2K]

Answer:

$u=\frac{9}{t}-\frac{1}{2} a t$