Which is equivalent to 4^3/4^6 ? A) 1/ 4^2 B) 4^-3 C) 1/64 D)1/4^3

3x+y=115x-y=21 Directions: Solve each system of equations by elimination. Clearly identify your solution.

olchik [2.2K]

To solve, we will follow the steps below:

3x+y=11 --------------------------(1)

5x-y=21 ------------------------------(2)

since y have the same coefficient, we can eliminate it directly by adding equation (1) and (2)

adding equation (1) and (2) will result;

8x =32

divide both-side of the equation by 8

x = 4

We move on to eliminate x and then solve for y

To eliminate x, we have to make sure the coefficient of the two equations are the same.

We can achieve this by multiplying through equation (1) by 5 and equation (2) by 3

The result will be;

15x + 5y = 55 ----------------------------(3)

15x - 3y =63 --------------------------------(4)

subtract equation (4) from equation(3)

8y = -8

divide both-side of the equation by 8

y = -1

8 0

1 year ago

One Step Equations solve for x 2.4+x=9.8

Simora [160]

Answer:

x=7.4

Step-by-step explanation:

2.4+x=9.8

(x=9.8-2.4)

x=7.4

7 0

3 years ago

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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:

$y_{p}(t) = C$

So

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

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

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:

$y_{p}(t) = C$

So

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

$(C)' + 4C = 8$

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

Lily is watering her grandmothers plants she is using a watering can that holds 58 gallon of water Lilly feels the watering can

Ipatiy [6.2K]

348 gallons! take 58 x 6

5 0

3 years ago

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What number in sequence of 11/14 19/28 4/7 13/28

laila [671]

13/28 < 4/7 < 19/28 < 11/14

4 0

3 years ago

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