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Dmitry_Shevchenko [17]
3 years ago
15

General terms of a sequence

Mathematics
1 answer:
MatroZZZ [7]3 years ago
5 0
(2n-1) /(n+9)
hope it can help u
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Help with this question please!!
notsponge [240]
18x can also be expressed as:

= 2x * (4+5)
= 2x * 9
= 18x

ANSWER: 2x * (4 + 5)

Hope this helps! :)
3 0
3 years ago
Read 2 more answers
David has been tasked with tracking the number of bagels sold (y) at freddy's 24-hour bagel shop from 12: 00 am (x= 0) to 12: 00
ICE Princess25 [194]

Step-by-step explanation:

let say he sold 5 bagels then he would sell 5 bagels

12-5=7 , b=7

replace in 12-5=7

b=12-s

8 0
3 years ago
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
2 years ago
help ASAP !!!!!!! A recipe calls for 2 cups of flour to make 24 cookies. What is the constant of proportionality that relates th
zavuch27 [327]

Answer:

1 cup of flour per 12 cookies

Step-by-step explanation:

2x=24y

divide by 2 on both sides

x=12y

5 0
3 years ago
PLEASE HELP!! GIVEN THE FOLLOWING TABLE OF DATA FROM A QUADRATIC EQUATION
antiseptic1488 [7]
The roots: this is when y=0, so in yours there are 2 roots. Just look at the x value when y=0 and that is your roots.

Y intercept- this is when x=0, so just look at the y value below x=0 and that is the y intercept. Note the answer will probably be in the form (0,_)

Vertex=do you see a pattern? Well the vertest would be the highest or lowest point of the quadratic equation. Your vertex would be (5,-9) because just look at x=4 and x=6, bit of the y values are -8 and when you look at x=3 and x=7 they are also the same because this is a quadratic equation.

Max or min: yours is a minimum because (5,-9) is the lowest point. Every value left and right of this are higher up the graph, so this would be a minimum.

*something that will help you see this all more clearly is if you graphed this or put it into Desmos to see the vertex etc.

7 0
2 years ago
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