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Vesnalui [34]
4 years ago
6

What can each term of the equation be multiplied by to eliminate the fractions before solving? x – + 2x = StartFraction one-half

EndFraction x minus StartFraction 5 Over 4 EndFraction plus 2 x equals StartFraction 5 Over 6 EndFraction plus x. + x 2 6 10 12
Mathematics
2 answers:
dalvyx [7]4 years ago
6 0

Answer: while solving an equation involving fractions we eliminate the fraction by multiplying the LCD of all the denominators present in the equation . LCD means Least common Denominator so for this question when we try to eliminate the denominator we first try to find the LCM (2,4,6) because that will give us the LCD.

2=2

4=2·2

6=2·3

LCM = 2·2·3

LCM = 12

It means we need to multiply the 12 to each term of equation to eliminate the fractions before solving.

12

irina [24]4 years ago
3 0

The answer to your question would be 12 or in alphabetical order D.  Hope this helps, have a nice day!

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Answer:

x=-3, y=-7. (-3, -7).

Step-by-step explanation:

4x+5=x-4

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The radius of a right circular cone is increasing at a rate of 1.1 in/s while its height is decreasing at a rate of 2.6 in/s. At
o-na [289]

Answer:

The volume of the cone is increasing at a rate of 1926 cubic inches per second.

Step-by-step explanation:

Volume of a right circular cone:

The volume of a right circular cone, with radius r and height h, is given by the following formula:

V = \frac{1}{3} \pi r^2h

Implicit derivation:

To solve this question, we have to apply implicit derivation, derivating the variables V, r and h with regard to t. So

\frac{dV}{dt} = \frac{1}{3}\left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right)

Radius is 107 in. and the height is 151 in.

This means that r = 107, h = 151

The radius of a right circular cone is increasing at a rate of 1.1 in/s while its height is decreasing at a rate of 2.6 in/s.

This means that \frac{dr}{dt} = 1.1, \frac{dh}{dt} = -2.6

At what rate is the volume of the cone changing when the radius is 107 in. and the height is 151 in.

This is \frac{dV}{dt}. So

\frac{dV}{dt} = \frac{1}{3}\left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right)

\frac{dV}{dt} = \frac{1}{3}(2(107)(151)(1.1) + (107)^2(-2.6))

\frac{dV}{dt} = \frac{2(107)(151)(1.1) - (107)^2(2.6)}{3}

\frac{dV}{dt} = 1926

Positive, so increasing.

The volume of the cone is increasing at a rate of 1926 cubic inches per second.

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Answer:

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