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2 years ago

7

A number is less than 30 but greater than 10.

1 answer:

GenaCL600 [577]2 years ago

7 0

The answer is 16 and 26. Both are less than 30 and greater than 10 and have 6 in the one’s place.

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Calculate the square root of 1/2

Oxana [17]

Answer:

I looked it up and it said 0.5

7 0

2 years ago

6x + 11 = 21<br> Help I’m on a quiz and this question is so confusing

Westkost [7]

Answer:

Step by Step Solution

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

6*x+11-(21)=0

Step by step solution :

STEP

1

:

Pulling out like terms

1.1 Pull out like factors :

6x - 10 = 2 • (3x - 5)

Equation at the end of step

1

:

STEP

2

:

Equations which are never true:

2.1 Solve : 2 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

2.2 Solve : 3x-5 = 0

Add 5 to both sides of the equation :

3x = 5

Divide both sides of the equation by 3:

x = 5/3 = 1.667

One solution was found :

x = 5/3 = 1.667

3 0

3 years ago

Which of the following equations describes the graph? Urgent lol

o-na [289]

Answer : I think 3rd is the answer

8 0

2 years ago

What is the measure of ZRMN?

Volgvan

Yes! Yup your right! Congrats yup!

3 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

2 years ago