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Marrrta [24]
2 years ago
7

Which of the following is a good way to quickly determine whether 3/12 and 1/4 are equal​

Mathematics
2 answers:
Bas_tet [7]2 years ago
5 0

For this case we must indicate a form to determine if the following expressions are equal:

\frac {3} {12}\ and\ \frac {1} {4}

It is observed that if we multiply the numerator and denominator of the second fraction by "3" we obtain the first fraction:

\frac {1 * 3} {4 * 3} = \frac {3} {12}

Similarly, if we divide the numerator and denominator of the first fraction between "3" we obtain the second fraction.

\frac {\frac {3} {3}} {\frac {12} {3}} = \frac {1} {4}

ANswer:

Divide the numerator and denominator of the first fraction between "3"  or

Multiply the numerator and denominator of the second fraction by "3"

Novay_Z [31]2 years ago
4 0

Answer: Divide 3 into the numerator and denominator

Step-by-step explanation:

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A circle's radius that has an initial radius of 0 cm is increasing at a constant rate of 7 cm per second. Write a formula to exp
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Answer:

r = 7t

or not mathematically,

r(t) = 7t

r(t) means, r, which is a function of t.

Step-by-step explanation:

Initial size of the radius = 0 cm, at t = 0 s

Rate of increase of the radius of the circle = 7 cm/s

dr/dt = 7

dr = 7 dt

∫ dr = 7 ∫ dt

r = 7t + C (C is the constant of integration)

At t = 0, r = 0,

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3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

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