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3 years ago

Find the least common denominator for these two rational expressions. 3/2n -2/5n

Mathematics

2 answers:

Zielflug [23.3K]3 years ago

8 0

Every rational expression can be written in infinitely many equvialent forms which is 3/3n-15/2.

zmey [24]3 years ago

3 0

For 2n and 5n the least common denominator will be 10n

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Need help with mean , median , mode question + a few more..

Andreas93 [3]

Answer:

Mean:

a) 4

b) 2

c) 2

d) 2

Median:

a) 4,2

b) 5,2

c) 5,1

d) 4,1

Mode:

a) 1

b) 1

c) 1

d) 1

Step-by-step explanation:

Mean:

a) 1+4+2+1 = 8/4=4

b)1+5+2+1 = 9/4= 2.25 (round to the nearest whole number = 2)

c) 1+5+1+0 = 7/4 = 1.75 (round to the nearest whole number = 2)

d) 1+4+1+0 = 6/4 = 1.5 (round to the nearest whole number = 2)

3 0

3 years ago

A local farm sells strawberries and blueberries. The ratio of the number of strawberries to blueberries it sells at 5:3. If it s

olganol [36]

Answer:

125

Step-by-step explanation:

The ratio is 5:3 so we divide the 75 by 3 to get 1 part

1 part = 75

75 x 5 = 125

125 strawberries were sold :)

3 0

2 years ago

Y is a differentiable function of x. Choose the alternative that is the derivative dy / dx.

murzikaleks [220]

Differentiating both sides of

$x^3-y^3=1$

with respect to x yields (using the chain rule)

$3x^2 - 3y^2 \dfrac{\mathrm dy}{\mathrm dx} = 0$

Solve for dy/dx :

$3x^2 - 3y^2 \dfrac{\mathrm dy}{\mathrm dx} = 0 \\\\ 3y^2\dfrac{\mathrm dy}{\mathrm dx} = 3x^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{3x^2}{3y^2} = \dfrac{x^2}{y^2}$

The answer is then D.

3 0

2 years ago

Please help me at this wit steps how did you did it

balu736 [363]

Answer:

5. LCM of 7 and 14: 14.

multiples of 7: 7, 14

multiples of 14: 14

LCM of 8 and 12: 24.

multiples of 8: 8, 16, 24

multiples of 12: 12, 24

Step-by-step explanation:

$.$

5 0

3 years ago

A conical vessels with base radius 5 cm and height 24cm is full of water this water is emptied into a cylindrical vessel of base

Aliun [14]

Answer:

$2\ cm$

Step-by-step explanation:

$We\ are\ given\ that,\\Radius\ of\ the\ conical\ vessel=5\ cm\\Height\ of\ the\ conical\ vessel=24\ cm\\Hence,\\As\ we\ know\ that,\\Volume\ of\ a\ cone=\frac{1}{3}\pi r^2h\\Hence,\\The\ volume\ of\ the\ conical\ vessel=\frac{1}{3}\pi (5)^224= \frac{1}{3}\pi 25*24=200\pi \\Hence,\\The\ volume\ of\ the\ conical\ vessel=The\ amount\ of\ water\ filled\ in\ it\\Hence,\\The\ amount\ of\ water\ filled\ in\ the\ conical\ vessel=200\pi\\Lets\ consider\ the\ cylindrical\ vessel:\\Volume\ of\ a\ cylinder=\pi r^2h$

$As\ we\ are\ not\ said\ that\ the\ cylindrical\ vessel\ was\ empty,\ let\ its\ initial\\ height\ be\ h_1\\Radius\ of\ the\ Cylindrical\ vessel=10\ cm\\Initial\ height\ of\ the\ cylindrical\ vessel=h_1\\Initial\ volume\ of\ the\ cylindrical\ vessel=\pi *10^2*h_1=100h_1\pi \\Initial\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi \\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=Final\ amount\ of\ water\\ in\ the\ cylindrical\ vessel +Amount\ of\ water\ in\ conical\ vessel$

$Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi +200\pi \\Now,\\Let\ the\ final\ height\ be\ h_2\\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=\pi *10^2*h_2=100h_2\pi \\Hence,\\100h_2\pi =100h_1\pi +200\pi \\Hence,\\100h_2=100h_1+200\\100h_2-100h_1=200\\100(h_2-h_1)=200\\h_2-h_1=2\\Hence,\\As\ the\ rise\ in\ level\ is\ the\ difference\ between\ the\ initial\ and\ final\\ heights:\\The\ rise\ in\ level\ of\ water=2\ cm$

8 0

3 years ago