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3 years ago

The distance traveled by Earth in one orbit around

Mathematics

1 answer:

Zolol [24]3 years ago

7 0

Answer:

Step-by-step explanation:

v=d/t

v=(580000000mi/y)(y/365d)(d/24h)=66210 mph (rounded to nearest mph)

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Explain how you can use base ten block to find 2.16÷3

Andrei [34K]

You can use base ten blocks to find 2.16/ 3 by using 7 base ten blocks and 2 base one blocks.

6 0

3 years ago

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Solve 5x^3 - 3x^2 - 14x

Pie

<h3>I’ll teach you how to solve 5x^3 - 3x^2 - 14x</h3>

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5x^3 - 3x^2 - 14x

Factor out the common term x:

x(5x^2 - 3x 14)

Factor 5x^2 - 3x 14:

Break the expression into groups:

(5x62+7x)+(-10x-14)

Factor out x from 5x^2+7x:

x(5x+7)

Factor out -2 from -10x-14:

-2(5x+7)

x(5x+7)-2(5x+7)

Factor out the common term 5x+7:

(5x+7)(x-2)

x(x+7)(x-2)

Your Answer Is x(x+7)(x-2)

Plz mark me as brainliest :)

7 0

4 years ago

(8x - 6) - (7x - 9)<br><br> I hope you're open to me asking more questions in the reply <3

Gekata [30.6K]

Answer:

x + 3

Step-by-step explanation:

Given

(8x - 6) - (7x - 9)

Distribute the first parenthesis by 1 and the second by - 1

= 8x - 6 - 7x + 9 ← collect like terms

= x + 3

5 0

3 years ago

Describe the translation of the following function.<br><br>Y = -1/3 |x + 4| - 2

katrin2010 [14]

Y equals negative one over three. absolute value of x plus 4. minus 2

5 0

4 years ago

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Sum to n terms of each of following series. (a) 1 - 7a + 13a ^ 2 - 19a ^ 3+...

julia-pushkina [17]

Notice that the difference in the absolute values of consecutive coefficients is constant:

|-7| - 1 = 6

13 - |-7| = 6

|-19| - 13 = 6

and so on. This means the coefficients in the given series

$\displaystyle \sum_{i=1}^\infty c_i a^{i-1} = \sum_{i=1}^\infty |c_i| (-a)^{i-1} = 1 - 7a + 13a^2 - 19a^3 + \cdots$

occur in arithmetic progression; in particular, we have first value $c_1 = 1$ and for $n>1$ , $|c_i|=|c_{i-1}|+6$ . Solving this recurrence, we end up with

$|c_i| = |c_1| + 6(i-1) \implies |c_i| = 6i - 5$

So, the sum to $n$ terms of this series is

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \underbrace{\sum_{i=1}^n i (-a)^{i-1}}_{S'} - 5 \underbrace{\sum_{i=1}^n (-a)^{i-1}}_S$

The second sum $S$ is a standard geometric series, which is easy to compute:

$S = 1 - a + a^2 - a^3 + \cdots + (-a)^{n-1}$

Multiply both sides by $-a$ :

$-aS = -a + a^2 - a^3 + a^4 - \cdots + (-a)^n$

Subtract this from $S$ to eliminate the intermediate terms to end up with

$S - (-aS) = 1 - (-a)^n \implies (1-(-a)) S = 1 - (-a)^n \implies S = \dfrac{1 - (-a)^n}{1 + a}$

The first sum $S'$ can be handled with simple algebraic manipulation.

$S' = \displaystyle \sum_{i=1}^n i (-a)^{i-1}$

$\displaystyle S' = \sum_{i=0}^{n-1} (i+1) (-a)^i$

$\displaystyle S' = \sum_{i=0}^{n-1} i (-a)^i + \sum_{i=0}^{n-1} (-a)^i$

$\displaystyle S' = \sum_{i=1}^{n-1} i (-a)^i + \sum_{i=1}^n (-a)^{i-1}$

$\displaystyle S' = \sum_{i=1}^n i (-a)^i - n (-a)^n + S$

$\displaystyle S' = -a \sum_{i=1}^n i (-a)^{i-1} - n (-a)^n + S$

$\displaystyle S' = -a S' - n (-a)^n + \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle (1 + a) S' = \dfrac{1 - (-a)^n - n (1 + a) (-a)^n}{1 + a}$

$\displaystyle S' = \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2}$

Putting everything together, we have

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 S' - 5 S$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2} - 5 \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} =\boxed{\dfrac{1 - 5a - (6n+1) (-a)^n + (6n-5) (-a)^{n+1}}{(1+a)^2}}$

8 0

2 years ago