15-g=23-2g math please help

Please help!!!!<br><br>what is the least common denominator of 1/6, 10/11, 5/12?<br>

Vesna [10]

<h3>Answer: LCM = 132</h3>

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Work Shown:

LCM = least common denominator

List out the prime factorization of each denominator

6 = 2*3
11 = 1*11
12 = 2*2*3

So we have the list of primes 2,3, and 11 that help form the denominators when we multiply some of them together.

The prime 2 shows up at most twice, so 2*2 = 4 is a factor of the LCM

The prime 3 shows up at most one time, meaning 3 is also a factor

The prime 11 shows up at most one time, so 11 is another factor

Multiply these factors to get 4*3*11 = 12*11 = 132

The LCM is 132

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Another Approach:

Focus on 1/6 and 10/11 for now. The LCM is 66 because 6*11 = 66. We simply multiply the denominators together. Then we divide over the GCF 1 to get 66/1 = 66.

The LCM of 1/6 and 10/11 is 66

The fractions 1/6 and 10/11 are equivalent to 11/66 and 60/66 respectively

The original list of fractions updates to 11/66, 60/66, 5/12

We've gone from 3 different denominators to now 2 different denominators.

Repeat the steps of multiplying the denominators and dividing by the GCF

66*12 = 792

792/(gcf of 66 and 12) = 792/6 = 132

So the LCM of all the fractions is 132.

3 0

3 years ago

1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat

brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

$x = 2$ is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

5 0

2 years ago

Make y the subject to the following questions

skelet666 [1.2K]

In this question it basically wants you to leave Y alone in a side of the equation.

In this case,
For 3y=c

$y=\frac{c}{3}$

For Ay=w

$y=\frac{w}{A}$

For Y/c=w
Y=cw

For y/a=2c
y=2ac

For a=y+p
y=a-p

For C=y-k
y=C+k

5 0

2 years ago

Find the 10th term of the geometric sequence 1, 3, 9, .

spin [16.1K]

Answer:

19683

Step-by-step explanation:

As the formula of the geometric sequence:

$a_{n} =a_{1} *r^{n-1}$