The sum of two numbers is 55. One number is 4 times as large as the other. What are the numbers? PLEASE HELP

Emily has pennies and five dollar bills in her pocket. The number of pennies is eight less than three times the number of five d

Zina [86]

Answer:

3f - 8 = number of pennies

Step-by-step explanation:

3f - 8 = number of pennies

6 0

3 years ago

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3/15 + 9/15 + 26/15=

klasskru [66]

Answer:

Mixed number - 2 8/15

decimal - 2.53

Step-by-step explanation:

3+9+26=38

denominator is 15

38/15 convert to a mixed number =

2 8/15 as a decimal is 2.53

4 0

4 years ago

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The radius of the circle is 3 millimeters. what is the circles area

Irina18 [472]

Answer:

28.26 millimeters

4 0

3 years ago

PLEASE HELP ME!!!!

Nat2105 [25]

8 * 2 * 10 = 160 * 10 = 1600

7 0

3 years ago

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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

3 years ago