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Taya2010 [7]
3 years ago
11

Find the 3rd term in the sequence

Mathematics
1 answer:
klemol [59]3 years ago
4 0

Step-by-step explanation:

a(n) = a ( n - 1) - 17

a(1) = 20

a(2) = a ( 2 - 1) - 17

= a (1) - 17

= 20 - 17

= 3

Now

a(3) = a ( 3 - 1 ) - 17

= a (2) - 17

= 3 - 17

= - 14

Hope it will help you.

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What’s the solution set for x? Can someone please answer and explain how they got they’re answer for 40 points.
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3 0
2 years ago
Jameson has 4300 in credit card debt with 14% interest that he wants to pay off in 24 months. He will need to make monthly payme
lakkis [162]

Answer:

Total cost of repayment  = 4955.04

Net interest paid = 655.04

Step-by-step explanation:

Given

Amount taken on loan = 4300

Repayment plan

Monthly installment= 206.46

Yearly installment  = 206.46 * 12 = 2477.52

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Time Period for repaying the loan = 2 years

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= 206.46 *24\\= 4955.04

Net interest paid

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Net interest paid = 655.04

8 0
2 years ago
How many solutions does the following equation have? -3z + 9 -2z = -12 - 5z
asambeis [7]
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4 0
3 years ago
Use the Binomial Theorem/Pascal's Triangle to expand (2a + 2b)^5
pochemuha

Answer:  

32a^5+160a^4b+320a^3b^2+320a^2b^3+160ab^4+32b^5

============================================================

Explanation:

Let's use Pascal's Triangle

In the row that starts with 1,5,... we have the values 1,5,10,10,5,1

These will be the coefficients of the terms.

Let x = 2a and y = 2b

We want to expand out (x+y)^5

Using pascals triangle, we get the following expansion

(x+y)^5 = 1x^5y^0+5x^4y^1+10x^3y^2+10x^2y^3+5x^1y^4+1x^0y^5

The numbers in bold are the coefficients 1,5,10,10,5,1 found earlier.

Note how the exponents for x start at 5 and count down to 0; while the y exponents start at 0 and count up to 5. For any term, the x and y exponents always add to 5 for the expansion of (x+y)^5. In general, the exponents of any term will add to n for (x+y)^n.

At this point, we plug in x = 2a and y = 2b

Since this will clutter things a bit, I'll do it term by term

  • 1x^5y^0 = 1(2a)^5(2b)^0 = 1(32a^5)(1) = 32a^5
  • 5x^4y^1 = 5(2a)^4(2b)^1 = 5(16a^4)(2b) = 160a^4b
  • 10x^3y^2 = 10(2a)^3(2b)^2 = 10(8a^3)(4b^2) = 320a^3b^2
  • 10x^2y^3 = 10(2a)^2(2b)^3 = 10(4a^2)(8b^3) = 320a^2b^3
  • 5x^1y^4 = 5(2a)^1(2b)^4 = 5(2a)(16b^4) = 160ab^4
  • 1x^0y^5 = 1(2a)^0(2b)^5 = 1(1)(32b^5) = 32b^5

So in the end, the expression (2a+2b)^5 expands out to

32a^5+160a^4b+320a^3b^2+320a^2b^3+160ab^4+32b^5

The binomial theorem uses the same basic idea, but instead of using Pascal's Triangle to get the coefficients, you'll use the nCr combination formula.

8 0
2 years ago
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