Ask question

Ask question

All categories

3 years ago

7

What is y={-\dfrac{1}{3}}x-9y=− 3 1 x−9y, equals, minus, start fraction, 1, divided by, 3, end fraction, x, minus, 9 written i

n standard form?

1 answer:

Kobotan [32]3 years ago

5 0

Answer:

$x+3y=-27$

Step-by-step explanation:

We are given that

$y=(-\frac{1}{3})x-9$

We have to find the standard form of given equation

$y=\frac{-x-27}{3}$

$3y=-x-27$

By using multiplication property of equality

$x+3y=-27$

We know that

Standard form of equation

$ax+by=c$

Therefore, the standard form of given equation is given by

$x+3y=-27$

You might be interested in

27. Is the following relation a function?

Sunny_sXe [5.5K]

Answer:

No

Step-by-step explanation:

27 No

28.A

you are to read it maximum and minimum point on the vertical axis

8 0

3 years ago

Read 2 more answers

Plastic bags used for packaging produce are manufactured so that the breaking strength of the bag is normally distributed with a

klasskru [66]

The answer is 2/6 in the fourth decimal place

4 0

3 years ago

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

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

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

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

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

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

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

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

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

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

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

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

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

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

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

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

Another payment type enabled by smartphones is person-to-person payments, which are apps that allow people to send money electro

Galina-37 [17]

Answer and Step-by-step explanation:

Person to person payment is an online technology that allows customers to transfer their money and funds from their account to another account through mobile phones. A p2p app allows one user to send money to another user by using an app or website—transaction including anything from paying a dinner bill, rent, or contributing to charity.

The increased obtaining of online banking, mobile banking, and e-commerce by users has paved the way for greater use of person-to-person payments. In the payment markets sending money between smartphones has become an ordinary matter. According to Billtrust, young adults are using a person to person payment more than two generations.

Nearly half of smartphone owners regularly used p2p payment apps. Most people used p2p payment app because it offers better security or most of their peers use it.

Millennials have often led older Americans in their adoption and use of technology. More than 93% millennial (who turns ages 23 to 38 this year) have their smartphones. Similarly, the vast majority of millennials use social media, compared with smaller shares among old generations.

3 0

3 years ago

Plz someone help me I just guessed but what's the real answer, I need this asap

sladkih [1.3K]

Answer: Honestly it has to be B dont see how the others come into play

Step-by-step explanation:

6 0

3 years ago