Ask question

Ask question

All categories

elena-14-01-66 [18.8K]

3 years ago

10

Please help!!!!! I’ll give you the top comment thing or what is called

1 answer:

VashaNatasha [74]3 years ago

5 0

$2,250 would be the answer

1100

+500

______

1600+(100×2)=1800

+150 +100=2050+50=2100+150=2250

You might be interested in

Help !! I can’t find the answer

jeka94

Answer:

385/pi

Step-by-step explanation:

Circumference is given by

C= pi * d where d is the diameter

385 = pi *d

Divide each side by pi

385/pi = pi * d/pi

385/pi = d

7 0

3 years ago

This??? What is wrong with it?

ELEN [110]

Answer:

15.8 sq. in. of paper will be required.

Step-by-step explanation:

The problem is that a drinking cup does not have a cover, so only the lateral surface area counts.

I.e. We need only the first term.

A = pi r l = pi * 1.5 * sqrt(3^2+1.5^2)

= 15.81 sq. in.

4 0

3 years ago

Math Plz Help. <br><br>Please graph with two points! <br>explain how you found your answer

Softa [21]

X intercept: y = 0
y= -3/4x -4
0 = -3/4 x - 4
4= -3/4 x
4 • -4/3 = x
-5.3 = x
y intercept: x = 0
y = -3/4 x -4
y = -3/4 (0) -4
y = -4
the points are (-5.3, 0) and (0, -4)

3 0

3 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

I need help with this

Softa [21]

Nothing can be done with this question!

8 0

3 years ago