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3 years ago

Find the smallest value of k such that the LCM of k and 6 is 60

Mathematics

1 answer:

Advocard [28]3 years ago

5 0

Answer: The smallest valuest value for<em> k </em>is 10, such that LCM o<em>f k</em> and 6 is 60.

Step-by-step explanation:

We know that, LCM = Least common multiple.

For example : LACM of 12 and 60 is 60.

If LCM of k and 6 is 60.

i.e. the least common multiple of k and 6 is 60.

Since, 10 x 6 = 60.

The smallest valuest value for<em> k </em>should be 10, such that LCM o<em>f k</em> and 6 is 60.

Hence, the smallest value of k is 10.

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3 years ago

Which is a simplified form of the expression 5x – 12 + 18x – 4?

Veseljchak [2.6K]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ You may have forgotten to add the options

However, you can collect like terms

In other words, collect the 'x' values and collect the numbers

5x + 18x = 23x

-12 - 4 = -16

Thus, your answer would be 23x - 16.

Any further questions, let me know.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

4 0

3 years ago

Read 2 more answers

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

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3 years ago

Which amount is a reasonable tip for an airport skycap? $2 total $2 per bag $5 per bag $10 per bag

Allushta [10]

Hey there!

Every bag is $2

Hope I helped!!

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3 years ago

A motorboat can go 16 miles downstream on a river in 20 minutes. It takes 30 minutes for this boat to go back upstream the same

Katyanochek1 [597]

Answer:

A) d = 48t, d in miles, t in hours

B)d = 32t, d in miles, t in hours

C)y = 8 mph

Step-by-step explanation:

16 miles in 20 mins = 48 mph

16 miles in 30 mins = 32 mph

The boat speed is the average = (48 + 32)/2 = 40 mph

The current is the difference = 8 mph

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3 years ago