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3 years ago

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This is due Monday but I want to get it done today, because the teacher is still going to check it, and I have practicly nothing

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1 answer:

Crank3 years ago

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Let f(x)=15/(1+4e)^(-0.2x)<br><br> What is the y-intercept of the graph of f(x) ?

sergejj [24]

That means when x=0
e^-0.2x =e^0=1
So f(0)=15/5=3

5 0

3 years ago

Read 2 more answers

Sam has a rectangular rug with an area of 50 square feet . The rug is 5 feet longer than it is wide.

ivolga24 [154]

Answer:

4 feet

Step-by-step explanation:

that's the right answer

3 0

2 years ago

The number of girls in a classroom is greater than the number of boys. There are more than girls. Write system of inequalities t

Fofino [41]

Answer:

Step-by-step explanation:

Required

Represent the relationship between population of girls and boys

From the question, we have that

$Boys = x\\Girls = y$

<em>Given that y is greater than x in the classroom;</em>

<em>One of the keywords in the given data is</em><em> "Greater than"</em>

The inequality to represent the system is the greater than sign;

In other words, >

Bringing x, y and > together to form a complete inequality; we have

$Girls > Boys$

Replace Girls with y and Boys with x

$y > x$

5 0

2 years ago

Melissa is making clothes for her dolls she has 7/8 yard of fabric each doll shirts require 2/7 of a yard of fabric. how many sh

vichka [17]

7/8 / 2/7 =

7/8 * 7/2 = 49/16 = 3 1/16

she can make 3 shirts

4 0

2 years ago

NO LINKS OR FILES!

Archy [21]

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

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

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

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

(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 <em>t</em> 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 <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √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 <em>v(t)</em> is the derivative of <em>s(t)</em>, 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