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fgiga [73]
3 years ago
9

Need help fast Thank

Mathematics
1 answer:
ollegr [7]3 years ago
3 0
What you need help with
You might be interested in
Help please, 20 points!
Veronika [31]
The answer is B)\sqrt[(-15, 4)]{544}

Hopes this helps ya!
7 0
3 years ago
How do you do this question?
Norma-Jean [14]

Answer:

Divergent

Step-by-step explanation:

an = 3 sin(2/n)

Choose bn = 3 (2/n).

Using Limit Comparison Test:

lim(n→∞) an / bn

= lim(n→∞) [3 sin(2/n)] / [3 (2/n)]

= lim(n→∞) [sin(2/n)] / (2/n)

= 1

The limit is greater than 0, and bn diverges, so an also diverges.

5 0
4 years ago
A carpenter bought 5 identical boxes of nails. She used 25 nails fir a project and now has 435 nails left. Which equation can be
stiks02 [169]

Answer:

The equation that can be solved to find the number of nails in each box =

5x = 460

The number of nails in each box = 92 nails

Step-by-step explanation:

A carpenter bought 5 identical boxes of nails. She used 25 nails for a project and now has 435 nails left. Which equation can be solved to find the number of nails in each box.

Let the number of nails in each box be represented by x

Therefore our Equation =

5x = 25 + 435

5x = 460

Solving for x

5x/5 = 460/5

x = 92 nails

Therefore, the number of nails in each box = 92 nails

5 0
3 years ago
How many values are in the range 17 to 118? <br><br> A- 102<br> B- 135<br> C- 100<br> D- 67.5
Tasya [4]

Answer:

C- 100 values

Step-by-step explanation:

3 0
2 years ago
Find the first four terms of the recursive sequence: a1 = 8 and an = an-1+n.
DedPeter [7]

Answer:

a1 = 8

a2 =10

a3 = 13

a4 = 17

Step-by-step explanation:

The first term is a1 = 8

The second term is a2 = a1 + n

                                       = 8+2

                                       = 10

The third term is a3 = a2 + n

                                       = 10+3

                                       = 13

The fourth term is a4 = a3 + n

                                       = 13+4

                                       = 17

8 0
4 years ago
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