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Dimas [21]
2 years ago
6

Algebra 2 math. Can someone help me please?

Mathematics
1 answer:
bogdanovich [222]2 years ago
5 0

Answer:

wruwgwjwggwwfjjwtjgwjwgjwjyjwgek

no one help you this is bad app hahahahaha

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A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 ​min, t
ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

T(t)=T_a+(T_0-T_a)e^{-kt}

where T_a is the ambient temperature, T_0 is the initial temperature, t is the time and k is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say T_0

We know that the ambient temperature is T_a=65, so

T(t)=65+(T_0-65)e^{-kt}

We also know that when t=5 \:min the temperature is T(5)=35 and when t=10 \:min the temperature is T(10)=50 which gives:

T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}

T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}

Rearranging,

35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}

50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}

If we divide these two equations we get

\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}

\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}

Now, that we know the value of k we can use it to find the initial temperature of the beam,

35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5

so the beam started out at 5 °F.

6 0
3 years ago
Prove or give a counterexample:
natima [27]

Answer: Although A and AT have the same eigenvalues, they may actually have different eigenvectors for that particular eigenvalue. Therefore A and AT have at least one eigenvector that aren't in common.

Step-by-step explanation:

These two attachment explains the provings and counterexamples briefly and simply

6 0
3 years ago
Two solutions to y'' – 2y' – 35y = 0 are yı = e, Y2 = e -5t a) Find the Wronskian. W = 0 Preview b) Find the solution satisfying
pashok25 [27]

Answer:

a.w(t)=-12e^{2t}

b.y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}

Step-by-step explanation:

We have a differential equation

y''-2 y'-35 y=0

Auxillary equation

(D^2-2D-35)=0

By factorization method we are  finding the solution

D^2-7D+5D-35=0

(D-7)(D+5)=0

Substitute each factor equal to zero

D-7=0  and D+5=0

D=7  and D=-5

Therefore ,

General solution is

y(x)=C_1e^{7t}+C_2e^{-5t}

Let y_1=e^{7t} \;and \;y_2=e^{-5t}

We have to find Wronskian

w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}

Substitute values then we get

w(t)=\begin{vmatrix}e^{7t}&e^{-5t}\\7e^{7t}&-5e^{-5t}\end{vmatrix}

w(t)=-5e^{7t}\cdot e^{-5t}-7e^{7t}\cdot e^{-5t}=-5e^{7t-5t}-7e^[7t-5t}

w(t)=-5e^{2t}-7e^{2t}=-12e^{2t}

a.w(t)=-12e^{2t}

We are given that y(0)=-7 and y'(0)=23

Substitute the value in general solution the we get

y(0)=C_1+C_2

C_1+C_2=-7....(equation I)

y'(t)=7C_1e^{7t}-5C_2e^{-5t}

y'(0)=7C_1-5C_2

7C_1-5C_2=23......(equation II)

Equation I is multiply by 5 then we subtract equation II from equation I

Using elimination method we eliminateC_1

Then we get C_2=-\frac{5}{2}

Substitute the value of C_2 in  I equation then we get

C_1-\frac{5}{2}=-7

C_1=-7+\frac{5}{2}=\frac{-14+5}{2}=-\frac{9}{2}

Hence, the general solution is

b.y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}

7 0
3 years ago
Alguien me ayuda en esto
guajiro [1.7K]

Answer:

diagrama c

Step-by-step explanation:

5 0
3 years ago
A storage room measures 48 inches by 60 inches.what is the total area in square feet of the closet?
Mandarinka [93]
2880 square feet squared
7 0
3 years ago
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