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4 years ago

Como encontrar el valor de la X

Mathematics

1 answer:

Anarel [89]4 years ago

7 0

Can't understand your langrage

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Julli [10]

Add dress and shoes together:

30 + 25 = 55

1/5 off means the total after discount would be 4/5 ( 1 - 1/5 = 4/5)

Multiply total price by 4/5:

55 x 4/5 = (55 x4)/5 = 220/5 = 44

Total after discount is 44

Now add shipping:

44 + 8 = 52

Total paid = £52

7 0

3 years ago

Read 2 more answers

A line with (5,6) and a slope of 2/15

BlackZzzverrR [31]

Slope, shmlope. What's the q?

4 0

4 years ago

19. answer this question.

Irina-Kira [14]

<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ final = original x multiplier^n

n is the number of years

Substitute the values in:

final = 3800 x 1.054^3

final = 4449.440763

The closest answer is A

➶ Hope This Helps You!

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5 0

4 years ago

Read 2 more answers

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago

How many solutions does the system have?

serious [3.7K]

The system has two solutions.

You can tell because the second equation is a quadratic and the first is a line, so the line passes through the quadratic twice.

8 0

3 years ago