Simplify 1 17/12 Explain

What kind of decimal is 0.05? How do you know?

solniwko [45]

Answer:

0.05 is a terminating decimal

Step-by-step explanation:

0.05 is a terminating decimal because it has finite digits.

3 0

2 years ago

Read 2 more answers

Justin, Cam, and Ben are playing a board game where exactly one player will win. Ben estimates that Justin has a 20\%20%20, perc

motikmotik

Answer:

false

Step-by-step explanation:

Based on the information provided we can say that this is false. Since it is a board game and no actual information regarding each players position in the game has been presented, then each player has an equal chance of winning each game. Therefore since there is a total of 3 players the percent chance of winning each game for each player is 33% (100 / 3 = 33)

6 0

3 years ago

Help me with my test

pychu [463]

Answer:

A

Step-by-step explanation:

5 0

2 years ago

Sarah is purchasing pencils to share. Each package has 12 pencils. The equation n = 12p, where n is the total number of pencils

podryga [215]

Answer and Step-by-step explanation:

Each package has 12 pencils

total number of pencils Sarah purchased: n = 12p

n = total number of pencils

p = number of packages

n depends on p, because the total number of pencils dependos on how many packages she will buy, so:

independent variable: p

dependent variable: n

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║ p ║ n = 12p ║ n ║

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║ 3 ║ n = 12*3 ║ 36 ║

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║ 4 ║ n = 12*4 ║ 48 ║

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║ 5 ║ n = 12*5 ║ 60 ║

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║ 6 ║ n = 12*6 ║ 72 ║

╠═══╬══════════╬════╣

║ 7 ║ n = 12*7 ║ 84 ║

╚═══╩══════════╩════╝

7 0

3 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago