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charle [14.2K]
3 years ago
6

Compute the value of each expression: −4|−3−3|

Mathematics
2 answers:
alexandr1967 [171]3 years ago
8 0

Answer:-24

Step-by-step explanation:

zvonat [6]3 years ago
6 0

Answer:

-24

Step-by-step explanation:

−4|−3−3|

First find the value inside the absolute value

−4|-6|

Absolute value means take the non-negative value

-4 * 6

Multiply

-24

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Answer:

1). \frac{2}{x^{2}-x-12 }=\frac{2}{(x+3)(x-4)}

2). \frac{1}{x^{2}-16 }=\frac{1}{(x-4)(x+4)}

Step-by-step explanation:

In this question we have to write the fractions in the factored form.

Rational expressions are \frac{2}{x^{2}-x-12 } and \frac{1}{x^{2}-16 }.

1). \frac{2}{x^{2}-x-12 }

Factored form of the denominator (x² - x - 12) = x² - 4x + 3x - 12

                                                                           = x(x - 4) + 3(x - 4)

                                                                           = (x + 3)(x - 4)

Therefore. \frac{2}{x^{2}-x-12 }=\frac{2}{(x+3)(x-4)}

2). \frac{1}{x^{2}-16 }

Factored form of the denominator (x² - 16) = (x - 4)(x + 4)

[Since (a²- b²) = (a - b)(a + b)]

Therefore, \frac{1}{x^{2}-16 }=\frac{1}{(x-4)(x+4)}

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g Use this to find the equation of the tangent line to the parabola y = 2 x 2 − 7 x + 6 at the point ( 4 , 10 ) . The equation o
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Answer:

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}

y'=4x-7+0               Simplifying and apply constant rule: (c)'=0

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.

6 0
3 years ago
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