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skad [1K]
3 years ago
9

Solve the integral:

20dx" id="TexFormula1" title=" \int \: {e}^{ - {x}^{2} } dx" alt=" \int \: {e}^{ - {x}^{2} } dx" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Sever21 [200]3 years ago
8 0

e^{-x^2} has no antiderivative in terms of elementary functions (polynomials, exponentials, logarithms, trigonometric functions, etc), but there is a special function defined to fit that role called the error function, \mathrm{erf}(x), where

\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt

By the fundamental theorem of calculus, we can see that

\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}

which means we have

\displaystyle\int e^{-x^2}\,\mathrm dx=\dfrac{\sqrt\pi}2\mathrm{erf}(x)+C

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You use 4 gallons of water on 30 plants in your garden. At that rate, how much water will it take to water 45 plants?
OverLord2011 [107]
Question that needs to be solved:You use 4 gallons of water on 30 plants in your garden. At that rate, how much water will it take to water 45 plants?=> 4 gallons is to 30 plants.=> 4 gallons / 30 plants = 0.13 gallons per plants.Now, it needs to water around 45 plants, let us solve how much will it need => 0.13 gallons * 45 plants = 6 gallons<span>Thus, to water 45 plants you need to have 6 gallons of water</span>
5 0
3 years ago
Lynette wanted to run one mile in eight minutes. To be on pace to reach
natta225 [31]

Answer:

0.25 miles

Step-by-step explanation:

assuming that she keeps a constant Velocity/ speed,

find the speed she intends to run with, ie total distance/time

\frac{1}{8} miles \: per \: minute

find the distance she travels in two minutes, ie speed multiplied by time.

\frac{1}{8}  \times 2 =  \frac{1}{4 }  = 0.25

4 0
3 years ago
Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1
algol [13]

Find the critical points of f(x,y):

\dfrac{\partial f}{\partial x}=-2x=0\implies x=0

\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12

All three points lie within D, and f takes on values of

\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}

Now check for extrema on the boundary of D. Convert to polar coordinates:

f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t

Find the critical points of g(t):

\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0

\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2

\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi

where n is any integer. There are some redundant critical points, so we'll just consider 0\le t< 2\pi, which gives

t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3

which gives values of

\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}

So altogether, f(x,y) has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0
3 years ago
makayla's local movie theater has a movie goer club that charges an annual registration fee of $25. however, movie tickets are d
KonstantinChe [14]
<span>C(m) = 6x + 25 would be your answer. </span>
6 0
3 years ago
Read 2 more answers
The function f(x)=-10(x)(x-4) represents the approximate height of a projectile launch on the ground into the air as a function
ELEN [110]

Answer:

4 seconds.

Step-by-step explanation:

The function f(x)=-10(x)(x-4) ........ (1), represents the approximate height of a projectile launch on the ground into the air as a function of time in seconds x.

Now, we are asked that for how long from the launch does the projectile stays in the air.

Therefore, we have to solve the equation (1) making f(x) as zero.

Hence, 10x(x - 4) = 0

⇒ x = 0 or x = 4

(As x can not be zero since at x = 0 sec, the projectile was at the ground.}

Hence, x = 4 seconds.

Therefore, the projectile was in the air for 4 seconds. (Answer)

4 0
3 years ago
Read 2 more answers
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