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2 years ago

Select the correct systems of equations.

Mathematics

1 answer:

fomenos2 years ago

4 0

Answer:

quads

Step-by-step explanation:

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Someone help me please

laila [671]

Answer:

37°

Step-by-step explanation:

You can see it is a right angle. Do 90-AOB angle=BOC angle.

4 0

2 years ago

4. A salesman has a basic salary and, in addition, recelves a commission which is a fixed percentage of his sales volume, when h

vekshin1

The commission is 20% of sales, the basic salary is $200 and the equation is y = 0.2x + 200

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Let y represent the total sales for x weekly sales, m represent the commision and b represent the basic salary. Hence:

400 = 1000m + b (1)

Also:

300 = 500m + b (2)

Hence:

m = 0.2 = 20%, b = $200

The commission is 20% of sales, the basic salary is $200 and the equation is y = 0.2x + 200

Find out more on equation at: brainly.com/question/2972832

8 0

2 years ago

Draw a number line that includes both positive and negative integers from +5 to -5.

aleksandrvk [35]

Answer:

5 4 3 2 1 0 -1 -2 -3 -4 -5

Step-by-step explanation:

6 0

2 years ago

Which value could go in the blank cell so that the percentage of only children that

babunello [35]

Answer:

maybe B

I am not to my answer

8 0

2 years ago

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

1 year ago