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Alla [95]
3 years ago
10

A package of 6 pairs of insulated socks costs $32.34. What is the unit price of the pairs of socks?

Mathematics
1 answer:
ANTONII [103]3 years ago
5 0
32.34/6 = $5.39 per sock
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Kryptonite is a material found on the planet Krypton and has various effects, most importantly on Superman. The most common type
Rashid [163]

Answer:

The maximum amount of red kryptonite present is 33.27 g after 4.93 hours.

Step-by-step explanation:

dy/dt = y(1/t - k)

separating the variables, we have

dy/y = (1/t - k)dt

dy/y = dt/t - kdt

integrating both sides, we have

∫dy/y = ∫dt/t - ∫kdt

㏑y = ㏑t - kt + C

㏑y - ㏑t = -kt + C

㏑(y/t) = -kt + C

taking exponents of both sides, we have

\frac{y}{t} = e^{-kt + C}  \\\frac{y}{t} = e^{-kt}e^{C} \\\frac{y}{t} = Ae^{-kt}   (A = e^{C})\\y = Ate^{-kt}

when t = 1 hour, y = 15 grams. So,

y = Ate^{-kt}\\15 = A(1)e^{-kX1}\\15 = Ae^{-k}(1)

when t = 3 hours, y = 30 grams. So,

y = Ate^{-kt}\\30 = A(3)e^{-kX3}\\30 = 3Ae^{-3k} (2)

dividing (2) by (1), we have

\frac{30}{15}  = \frac{3Ae^{-3k}}{Ae^{-k}} \\2 = 3e^{-2k}\\\frac{2}{3} = e^{-2k}

taking natural logarithm of both sides, we have

-2k = ㏑(2/3)

-2k = -0.4055

k = -0.4055/-2

k = 0.203

From (1)

A = 15e^{k} \\A = 15e^{0.203} \\A = 15 X 1.225\\A = 18.36

Substituting A and k into y, we have

y = 18.36te^{-0.203t}

The maximum value of y is obtained when dy/dt = 0

dy/dt = y(1/t - k) = 0

y(1/t - k) = 0

Since y ≠ 0, (1/t - k) = 0.

So, 1/t = k

t = 1/k

So, the maximum value of y is obtained when t = 1/k = 1/0.203 = 4.93 hours

y = 18.36(1/0.203)e^{-0.203t}\\y = \frac{18.36}{0.203}e^{-0.203X1/0.203}\\y = 90.44e^{-1}\\y = 90.44 X 0.3679\\y = 33.27 g

<u>So the maximum amount of red kryptonite present is 33.27 g after 4.93 hours.</u>

3 0
3 years ago
WILL MARK BRANLIEST TO ACCURATE ANSWER. URGENT ANSWER REQUIRED. ANSWER NOW.
ICE Princess25 [194]

Answer:

It’s D. 4

Step-by-step explanation:

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7 0
3 years ago
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interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm
Igoryamba

Answer:

The curvature is \kappa=1

The tangential component of acceleration is a_{\boldsymbol{T}}=0

The normal component of acceleration is a_{\boldsymbol{N}}=1 (2)^2=4

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}

where

\boldsymbol{T}} is the unit tangent vector.

\frac{ds}{dt}=|| \boldsymbol{r}'(t)}|| is the speed of the object

We need to find \boldsymbol{r}'(t), we know that \boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k} so

\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}

Next , we find the magnitude of derivative of the position vector

|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2

The unit tangent vector is defined by

\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}

\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)

We need to find the derivative of unit tangent vector

\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})

And the magnitude of the derivative of unit tangent vector is

||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2

The curvature is

\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1

The tangential component of acceleration is given by the formula

a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}

We know that \frac{ds}{dt}=|| \boldsymbol{r}'(t)}|| and ||\boldsymbol{r}'(t)}||=2

\frac{d}{dt}\left(2\right)\: = 0 so

a_{\boldsymbol{T}}=0

The normal component of acceleration is given by the formula

a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2

We know that \kappa=1 and \frac{ds}{dt}=2 so

a_{\boldsymbol{N}}=1 (2)^2=4

3 0
3 years ago
Amy wrote a check for $32 to pay her monthly electricity bill, but when balancing her checkbook, she accidentally recorded it as
ElenaW [278]
Her check register will have a surplus of 64 dollars compared to her actual balance. 
example
100 +32 = 132  (incorrect)
100 -32 = 68 (correct)
132 -68 = 64 surplus

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What is 25/30 in simplest form
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5/6 all you do is divide by 5
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