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3 years ago

Task 1:

Mathematics

1 answer:

laiz [17]3 years ago

4 0

Answer:

The toy's maximum height is $10.12\ \text{feet}$ .

Step-by-step explanation:

The equation that models the toy's height in feet from the ground at t seconds after he threw it is given by :

$h = -2t^2 + 7t + 4$ .....(1)

It is required to find the toy's maximum height. For maxima or minima, put

$\dfrac{dh}{dt}=0$

Plugging the value of t in above equation,

$\dfrac{d(-2t^2 + 7t + 4)}{dt}=0\\\\-4t+7=0\\\\t=\dfrac{7}{4}\\\\t=1.75\ s$

Now put t = 1.75 s in equation (1) such that,

$h = -2(1.75)^2 + 7(1.75) + 4\\\\h=10.12\ \text{feet}$

So, the toy's maximum height is $10.12\ \text{feet}$ .

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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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When writing numbers in base 10, we only allow coefficients of 0 through 9. Why is that?

Margaret [11]

Answer:

Because 0 counts as the first possible number. 0-9 gives you 10 possible numbers.

Step-by-step explanation:

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2 years ago

A deep sea diver descends below the surface of the water at a rate of 60 feet each minute. What is the depth of the diver after

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Answer:

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Consider a circle whose size can vary. The circumference of the circle is always 2π times as large as its radius. Let r represen

ser-zykov [4K]

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d) 5.89feet

Step-by-step explanation:

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3 years ago

NEED ANSWERED ASAP!!!!

ehidna [41]

Answer:

The answer is A

Step-by-step explanation:

f(x) = x+2

g(x) = x²

g(fx) = (x+2)²

= x²+4x+4

g(fx)(-5) = (-5)²+4(-5)+4

= 25 - 20 + 4

= 9

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