Ask question

Ask question

All categories

stealth61 [152]

3 years ago

10

A all falls off a cliff at a height of 256 feet if the equation for high as a fraction of time is h(t) =-16t2 + initial hey wher

e T is time in seconds and h(t) is high in feet, how many seconds will it take for the pedal to hit the ground?

1 answer:

yawa3891 [41]3 years ago

7 0

Answer:

t=4

Step-by-step explanation:

$h(t)=-16t^2+256\\\text{The pedal will hit the ground when h(t)=0.}\\h(t)=0,\ -16t^2+256=0\\16t^2=256, \ t^2=16, \ t=4 \ or -4.\\$

As t represents time, t=-4 can't be a solution.

You might be interested in

What is 1.9961 x 10^-7

kakasveta [241]

Answer:0.00000019961

Step-by-step explanation:

I moved the decimal place by the number of the exponent

4 0

3 years ago

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

3 years ago

Members of the Granola Gang were popping popcorn over the campfire.

nordsb [41]

Answ

JOSUE

Step-by-step explanation:

1 Y 2 3 4 5 6 7 8 9 10 BOOM ESO ES LA RESPUESTA

8 0

3 years ago

Solve the system using substitution:<br> y = -6x + 7<br><br> 3x – 2y = 16<br><br> helppp

bearhunter [10]

Answer:

x=-6 y=-1

Step-by-step explanation:

// Solve equation [2] for the variable y

[2] y = -x - 7

// Plug this in for variable y in equation [1]

[1] 3x - 2•(-x -7) = -16

[1] 5x = -30

// Solve equation [1] for the variable x

[1] 5x = - 30

[1] x = - 6

// By now we know this much :

x = -6

y = -x-7

// Use the x value to solve for y

y = -(-6)-7 = -1

Solution :

{x,y} = {-6,-1}

7 0

3 years ago

Read 2 more answers

What is 37% of 150$

lakkis [162]

Answer:

$55.50

Step-by-step explanation:

150 (0.37) = 55.5

8 0

3 years ago

Read 2 more answers