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3 years ago

Using the distributive property what will you do?

Mathematics

1 answer:

Art [367]3 years ago

5 0

Answer:

Rewrite expressions!

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Write the slope and y intercept of each equation:

vichka [17]

Answer:

A) slope = 2, y-intercept = -6

B) slope = -4, y-intercept = 6

Step-by-step explanation:

A) y = 2x - 6

The equation for a line is y = mx + b, where m represents the slope, and b represents the y-intercept. So, to find the slope and y-intercept, all we have to do is look at the line's equation.

Here, the m is 2, so the slope is 2

The b is -6, so the y-intercept is -6

B) y = -4x + 6

Here, the m is -4, so the slope is -4

The b is 6, so the slope is 6

You would graph the equation y = 4x + 6 by plotting a point at the y-intercept of the line, which would be 6. Then, for every time you move one space to the right, you'd plot a point four spaces up to show the slope of four.

7 0

3 years ago

2,5, 25/2, ... find the 10th term

notka56 [123]

Answer:

The $10^{th}$ term of the given sequence

$t_{10} = \frac{5^{9} }{2^{8} }$

Step-by-step explanation:

Step(i):-

Given sequence 2 , 5, $\frac{25}{2}$

First term a = 2

The difference of given geometric sequence

$d = \frac{r_{2} }{r_{1} } = \frac{5}{2}$

Step(ii):-

The $n^{th}$ term of the given sequence

$t_{n} = ar^{n-1}$

The $10^{th}$ term of the given sequence

$t_{10} = (2)(\frac{5}{2} )^{10-1}$

$t_{10} = (2)(\frac{5}{2} )^{9}= \frac{5^{9} }{2^{8} }$

8 0

3 years ago

What's the divisor for 4/12 =1/3

lozanna [386]

I'm gonna go with 1 (:

3 0

3 years ago

Read 2 more answers

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

The class with the greatest mean sales in a spring fundraiser will win a prize. The mean for Eric's class is $148. The mean for

bezimeni [28]

The reason it would in crese is because they combin their money

6 0

3 years ago