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3 years ago

Simplify the expression 5r + 8r

Mathematics

2 answers:

S_A_V [24]3 years ago

7 0

Answer:

13r

Step-by-step explanation:

Jlenok [28]3 years ago

5 0

Answer: 13r

Step-by-step explanation:

You just add 5r with 8r

and you'll get

13r

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Answer:

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Step-by-step explanation:

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b = 180 - 37 - 109 = 34 degrees (angles in a triangle sum to 180 degrees).

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2 years ago

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maria [59]

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2 years ago

HELP PLZZZZZ

Dmitry_Shevchenko [17]

Answer: Option (D) Add the exponents and keep the same base. Then find the reciprocal and change the sign of the exponent.

Explanation:

Given expression:

$(2^3)(2^{-4})$

Step-1: Add the exponents and keep the same base by using $x^a*x^b = x^{a+b}$ property.

$(2^3)(2^{-4}) = 2^{(3+(-4))} = 2^{-1}$

Step-2: The reciprocal of $2^{-1}$ is $\frac{1}{2^{-1}}$ , which is $2^1$

Step-3: The exponent of $2^1$ is +1; therefore, change the sign of exponent. The end expression will now become:

$2^{-1}$

Hence, the correct option is (D) Add the exponents and keep the same base. Then find the reciprocal and change the sign of the exponent.

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3 years ago

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2 years ago

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago

Simplify the expression 5r + 8r​

Simplify the expression 5r + 8r