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2 years ago

Lcm of 2 and 3 very very very urgent

Mathematics

2 answers:

Arturiano [62]2 years ago

6 0

$\huge\boxed{Hi!}$

What does LCM mean?

L stands for Least

C stands for Common

M stands for Multiple

Thus, LCM means "Least Common Multiple"

Now, what is the LCM of 2 and 3?

Let's list some of the multiples of both numbers:

2: 2, 4, 6, 8, 10, 12, 14..

3: 3, 6, 9, 12, 15...

$\huge\boxed{Answer:{\boxed{\sf{The\:LCM\:is\:6.}}}}$

Hope it helps! Enjoy your day!

~Just a determined gal

$\bf{erutaNtneliS$

Licemer1 [7]2 years ago

5 0

Answer: 6

Step-by-step explanation: 2x3=6 it’s the lowest multiple

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What is 33 over 16 in a decimal

TiliK225 [7]

Answer:

2.0625

Step-by-step explanation:

Convert the fraction to a decimal by dividing the numerator by the denominator.

4 0

3 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$