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3 years ago

What is the midpoint of the line segment with endpoints?

Mathematics

1 answer:

ella [17]3 years ago

6 0

Answer:

B. (1,2)

Step-by-step explanation:

Look at the image down below!! (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧

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Which decimal number is 17/21 equal to?

exis [7]

It is equal to 0.8 five star please?

8 0

3 years ago

Read 2 more answers

Find the image of triangle GHJ under the translation of (X,Y) ——> ( x-1, y-8)

MAVERICK [17]

Answer:

triangle KLM

Step-by-step explanation:

Take one point on the initial triangle and move it according to the given translation to locate the image. I used point G. Its coordinate plot is at

(2,3). I perform the operation of the given translation to this. (X1-X2, y1-y2).

So, (2-1, 3-8). The new coordinates will be (1,-5) and the only triangle with these coordinates is triangle KLM.

8 0

3 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

The proof that AMNS AQNS is shown.

FrozenT [24]

Answer:

The answer is "MS and QS".

Step-by-step explanation:

Given ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. we have to prove that ΔMNS ≅ ΔQNS.

As NR and MQ bisect each other at S

⇒ segments MS and SQ are therefore congruent by the definition of bisector i.e MS=SQ

In ΔMNS and ΔQNS

MN=QN (∵ MNQ is isosceles triangle)

∠NMS=∠NQS (∵ MNQ is isosceles triangle)

MS=SQ (Given)

By SAS rule, ΔMNS ≅ ΔQNS.

Hence, segments MS and SQ are therefore congruent by the definition of bisector.

The correct option is MS and QS

3 0

2 years ago

What is 14/26 simplified?

Reika [66]

14/26 simplified is 7/13

4 0

2 years ago