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3 years ago

Find the constant rate of change

Mathematics

1 answer:

777dan777 [17]3 years ago

4 0

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

and to get the slope in any linear relation all we need is two points, let me rewrite this table some $\begin{array}{|c|c|ll} \cline{1-2} x(h)&y(c)\\ \cline{1-2} 5&15\\ 8&24\\ 12&36\\ 24&72\\ \cline{1-2} \end{array}$

so from the table, let's use hmmm (8 , 24) and (24 , 72)

$(\stackrel{x_1}{8}~,~\stackrel{y_1}{24})\qquad (\stackrel{x_2}{24}~,~\stackrel{y_2}{72}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{72}-\stackrel{y1}{24}}}{\underset{run} {\underset{x_2}{24}-\underset{x_1}{8}}}\implies \cfrac{48}{16}\implies 3$

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Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

4 years ago

A bag of potatoes has a mass of 4 kilograms.Some potatoes are taken out.The mass is now 2305 grams.What is the mass of the potat

Colt1911 [192]

First, convert the kilograms to grams. One kilogram is 1000 grams. So multiply it by 1000. Then subtract 2305 from 4000. 4000-2305=1695. So 1695 grams were taken out.

5 0

3 years ago

Veronica has 7 yards of ribbon. Each bow requires 3/4 of a yard ribbon. How many bows will she be able to make with the ribbon s

Sonbull [250]

Answer && Step-by-step explanation:

7 yards == 4/4

by simple arithmetic remove 3/4 from 7 until you can no longer obtain 3/4 from it

7 => 6*1/4

6*1/4=>5*2/4

5*2/4=>4*3/4

4*3/4=>4

4=> 3*1/4

3*1/4=>2*2/4

2*2/4=>1*3/4

1*3/4=>1

1=>1/4

1/4 remains and 9 bows were made

8 0

3 years ago

Help me with this equation!

vlada-n [284]

The correct answer is B. -12

8 0

3 years ago

Read 2 more answers

the length of a rectangle is 3 feet more than twice the width. if the width is increased by four feet the perimeter of the new r

densk [106]

Answer:

The area of the original rectangle = 189 sq ft.

Step-by-step explanation:

Let, the width of the rectangle = m feet

So, the length of the rectangle = (2 m + 3) ft

Now, new width w' = (m + 4)

Perimeter of new rectangle (P') = 68 ft

Perimeter of a rectangle = 2 (LENGTH + WIDTH)

⇒68 ft = 2( L + W')

or, 68 ft = 2 [( 2m + 3) + (m + 4)]

or, 2 (3m + 7) = 68

or, 3m + 7 = 68/2 = 34

⇒ 3m = 34 - 7 = 27

⇒ m = 27/3 = 9

or, m = 9 ft

Hence, the original width of the rectangle = 9 ft

Original Length of the rectangle = 2m + 3 = 2(9) + 3 = 21 ft

Now, Area of the Rectangle = LENGTH X WIDTH

= 21 ft x 9 ft = 189 sq ft

Hence,the area of the original rectangle = 189 sq ft.

5 0

4 years ago