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3 years ago

Suppose you earned 8t-3 on monday and 6t+5 dollars on tuesday. what were your total earnings? simplify your answer

1 answer:

Nadusha1986 [10]3 years ago

6 0

Let
A=8t-3
B=6t+5

we know that
total earnings=A+B
so
total earnings=(8t-3)+(6t+5)------> (8t+6t)+(-3+5)------> (14t+2) dollars

the answer is
(14t+2) dollars

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y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

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$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

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Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

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Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

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$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

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jek_recluse [69]

Answer:

Is a consistent independent system

Step-by-step explanation:

we have

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$y=-x-7$ ----> equation A

$5x-5y=5$ -----> isolate the variable y

$y=x-5$ ---> equation B

Compare equation A and equation B

We can affirm that

The slopes are not equal ( so the lines are not parallel)

The lines are different

The product of their slopes is equal to -1 (the lines are perpendicular)

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