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Dovator [93]
3 years ago
7

What is the pattern of the following sequence? 6, 11, 16, 21, . . .

Mathematics
2 answers:
Furkat [3]3 years ago
7 0

6 + x = 11; x = 5

11 + x = 16; x = 5

Therefore the pattern is +5

Hope this helps!

vovikov84 [41]3 years ago
4 0

Answer: The pattern is plus 5 or +5

Step-by-step explanation:

6+5=11

11+5=16

16+5=21

And so on

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Use the calculator to find the product. (9. 1 × 109)(7. 5 × 102) What is the exponent of the power in scientific notation?.
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The exponent of the power is 11 in scientific notation.

<h3>Given </h3>

The numbers are;

The product of  (9. 1 × 10^9)(7. 5 × 10^2)

<h3>What is scientific notation?</h3>

Scientific notation is a method of expressing numbers in terms of a decimal number between 1 and 10, but not 10 itself multiplied by a power of 10.

To find the product of the given numbers follows all the steps given below.

The product of number is;

=(9.1 \times  10^9) \times (7.5 \times 10^2)\\\\= (9.1 \times 7.5) \times (10^{(9+2)})\\\\ =68.25 \times 10^{11}

Hence, the exponent of the power is 11 in scientific notation.

To know more about Scientific notation click the link given below.

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4 0
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

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
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