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Reil [10]
3 years ago
12

Please help ASAP for the brainliest answer

Mathematics
2 answers:
Wittaler [7]3 years ago
7 0

Answer:

arithmetic sequence

Step-by-step explanation:

The next number is 21

Ronch [10]3 years ago
5 0

Answer:

Triangular Number Sequence

21

Step-by-step explanation:

This sequence is called the Triangular Number Sequence.

To find the next term, add 6 to 15.

The next term in the sequence is 21.

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y=4

Step-by-step explanation:

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This is 8). the line is 5 on the y-axis.

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The table below represents the function F. 3 7 F(x) 4 15 5 31 7 127 8 255 The equation that represents this function is?​
timurjin [86]

The table is an exponential function, and the equation that represents this function is f(x) =2^x -1

<h3>The equation of the table</h3>

The table is given as:

x   F(x)

4    15

5    31

7    127

8    255

Rewrite the table as:

x   F(x)

4    16 - 1

5    32 - 1

7    128 - 1

8    256 - 1

Express as a base of 2

x   F(x)

4    2⁴ - 1

5    2⁵- 1

7    2⁷ - 1

8    2⁸ - 1

Notice that the function is 1 subtracted from 2 raised to the power of the input.

This is represented as:

f(x) =2^x -1

Hence, the equation of the table is f(x) =2^x -1

Read more about exponential functions at

brainly.com/question/9608962

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4 0
1 year ago
Graph the inequality.<br> 1
Mrrafil [7]

Answer:

hope it helps

Step-by-step explanation:

6 0
2 years ago
For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
Elina [12.6K]

Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

3 0
3 years ago
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