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3 years ago

8

Problem number 26 of the Rhind Papyrus says: Find a quantity such that when it is added to StartFraction 1 Over 4 EndFraction of

itself the result is a 15. The modern day equation that models this problem is x plus StartFraction 1 over 4 EndFraction x equals 15.. What is the solution to the equation? x = 10 x = 12 x = 15 x = 30 What is the fee Michael charges each new subscriber?

1 answer:

Rzqust [24]3 years ago

6 0

Answer:

x=12

Step-by-step explanation:

For the given situation , a quantity x is added to $\frac{1}{4} x$ gives 15.

We can set up equation as

$x+\frac{1}{4}x=15$

Multiply each term by 4 on both sides to get rid the denominator.

It gives,

4 x+x=60

Now, combine like terms

5 x=60

Divide both sides by 5

x=12.

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3 years ago

What is 0.012 in percent?

stepan [7]

Answer:

1.2%

Step-by-step explanation:

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2 years ago

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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.

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3 years ago

Find the average rate of change.<br><br> p(x) = 6x + 7 on [2, 2 + h] , h ≠ 0

siniylev [52]

Answer:

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Step-by-step explanation:

This is [p(2+h)) - p(2) ] / ((2 + h - 2)

= [ 6(2) + h) + 7 - (6(2) + 7)] / h

= ( 12 + 6h + 7 - 12 - 7) / h

= 6h / h

= 6.

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3 years ago