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Sati [7]
3 years ago
12

Please help me, you have to simply it

Mathematics
1 answer:
allsm [11]3 years ago
7 0

\bf \sqrt{720x^2y^3}~~ \begin{cases} 720=2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5\\ \qquad 2^2\cdot 2^2\cdot 3^2\cdot 5\\ \qquad (2^2)^2\cdot 3^2\cdot 5\\ \qquad 4^2\cdot 3^2\cdot 5\\ \qquad (4\cdot 3)^2\cdot 5\\ \qquad 12^2\cdot 5\\ y^3=y^{2+1}\\ \qquad y^2y \end{cases}\implies  \begin{array}{llll} \sqrt{12^2\cdot 5x^2y^2y} \\\\\\ 12xy\sqrt{5y} \end{array}

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Kulay, there is no such thing as a "step by step answer" here.  You seem to want a "step by step solution."

I must assume that by 4/5 you actually meant (4/5) and that by 2/3 you meant (2/3).  Then your equation becomes:

(4/5)w - 12 = (2/3)w. 

The LCD here is 5*3, or 15, so mult. every term by 15:

12w - 180 = 10w.   
Add 180 to both sides, obtaining 12 w - 180 + 180 = 10w + 180.     

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Marysya12 [62]

Answer:

y=-\frac{74}{21}x+452

Step-by-step explanation:

Given:

The level of a lake is falling linearly.

On Jan 1, the level is 452 inches

On Jan 21, the level is 378 inches.

Now, a linear function can be represented in the form:

y=mx+b

Where, 'm' is the rate of change and 'b' is initial level

x\to number\ of\ days\ passed\ since\ Jan\ 1

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So, let Jan 1 corresponds to the initial level and thus b = 452 in

Now, the rate of change is given as the ratio of the change in level of lake to the number of days passed.

So, from Jan 1 to Jan 21, the days passed is 21.

Change in level = Level on Jan 21 - Level on Jan 1

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Now, rate of change is given as:

m=\frac{-74}{21}

Hence, the function to represent the lake level is y=-\frac{74}{21}x+452

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3 years ago
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irina1246 [14]

Answer:

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1 and 3 look the same size and 2 looks the same as c (39°) so you jsut fill in for the letters

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