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MrRa [10]
3 years ago
12

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

Mathematics
1 answer:
Vesna [10]3 years ago
3 0
<h3>Answer: LCM = 132</h3>

===================================================

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

---------------------

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.

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The tangent line to the given curve at the given point is y=9x-26.

Step-by-step explanation:

To find the slope of the tangent line we to compute the derivative of y=2x^2-7x+6 and then evaluate it for x=4.

(y=2x^2-7x+6)'          Differentiate the equation.

(y)'=(2x^2-7x+6)'       Differentiate both sides.

y'=(2x^2)'-(7x)'+(6)'    Sum/Difference rule applied: (f(x)\pmg(x))'=f'(x)\pm g'(x)

y'=2(x^2)'-7(x)'+(6)'  Constant multiple rule applied: (cf)'=c(f)'

y'2(2x)-7(1)+(6)'        Applied power rule: (x^n)'=nx^{n-1}

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y'=4x-7                    Simplify.

Evaluate y' for x=4:

y'=4(4)-7

y'=16-7

y'=9 is the slope of the tangent line.

Point slope form of a line is:

y-y_1=m(x-x_1)

where m is the slope and (x_1,y_1) is a point on the line.

Insert 9 for m and (4,10) for (x_1,y_1):

y-10=9(x-4)

The intended form is y=mx+b which means we are going need to distribute and solve for y.

Distribute:

y-10=9x-36

Add 10 on both sides:

y=9x-26

The tangent line to the given curve at the given point is y=9x-26.

------------Formal Definition of Derivative----------------

The following limit will give us the derivative of the function f(x)=2x^2-7x+6 at x=4 (the slope of the tangent line at x=4):

\lim_{x \rightarrow 4}\frac{f(x)-f(4)}{x-4}

\lim_{x \rightarrow 4}\frac{2x^2-7x+6-10}{x-4}  We are given f(4)=10.

\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}

Let's see if we can factor the top so we can cancel a pair of common factors from top and bottom to get rid of the x-4 on bottom:

2x^2-7x-4=(x-4)(2x+1)

Let's check this with FOIL:

First: x(2x)=2x^2

Outer: x(1)=x

Inner: (-4)(2x)=-8x

Last: -4(1)=-4

---------------------------------Add!

2x^2-7x-4

So the numerator and the denominator do contain a common factor.

This means we have this so far in the simplifying of the above limit:

\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}

\lim_{x \rightarrow 4}\frac{(x-4)(2x+1)}{x-4}

\lim_{x \rightarrow 4}(2x+1)

Now we get to replace x with 4 since we have no division by 0 to worry about:

2(4)+1=8+1=9.

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