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Anton [14]
3 years ago
13

A sequence is defined by the recursive function

Mathematics
1 answer:
Shalnov [3]3 years ago
4 0

Answer:

100

Step-by-step explanation:

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What’s 5x[(26-4)-(4+6)
seropon [69]

Answer:

The Answer is: The factors reduce to 60x.

Step-by-step explanation:

First the parenthesis, then the bracket, then the multiplication:

5x[(26 - 4) - (4 + 6)]

5x[22 - 10]

5x[12]

60x

Hope this helps! Have an Awesome Day!! :-)

8 0
3 years ago
Read 2 more answers
0.075538 to 2 decimal place<br> 0.081770 to 3 significant figure
Anon25 [30]

Answer:0.08

0.0818

Step-by-step explanation:

6 0
2 years ago
I need help ASAP <br> Show your work: x - 8 = 12
choli [55]

Answer:

x = 20  

Step-by-step explanation:

x - 8 = 12                          Add 8 to both sides

x -8 + 8 = 12 + 8               Add -8 and 8; add 12 and 8

x = 20                      

6 0
2 years ago
Read 2 more answers
A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.
Verdich [7]

Answer:

2a - 3b + 4c = 1

Step-by-step explanation:

Given

a^2 + b^2 + c^2 = 2(a - b - c) - 3

Required

Determine 2a - 3b + 4c

a^2 + b^2 + c^2 = 2(a - b - c) - 3

Open bracket

a^2 + b^2 + c^2 = 2a - 2b - 2c - 3

Equate the equation to 0

a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0

Express 3 as 1 + 1 + 1

a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0

Collect like terms

a^2 - 2a + 1 + b^2 + 2b + 1 + c^2  + 2c + 1 = 0

Group each terms

(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

Factorize (starting with the first bracket)

(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

((a - 1)^2) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

((a - 1)^2) + (b^2 + b+b + 1) + (c^2  + 2c + 1) = 0

((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2  + 2c + 1) = 0

((a - 1)^2) + ((b + 1)(b + 1)) + (c^2  + 2c + 1) = 0

((a - 1)^2) + ((b + 1)^2) + (c^2  + 2c + 1) = 0

((a - 1)^2) + ((b + 1)^2) + (c^2  + c+c + 1) = 0

((a - 1)^2) + ((b + 1)^2) + (c(c  + 1)+1(c + 1)) = 0

((a - 1)^2) + ((b + 1)^2) + ((c  + 1)(c + 1)) = 0

((a - 1)^2) + ((b + 1)^2) + ((c  + 1)^2) = 0

Express 0 as 0 + 0 + 0

(a - 1)^2 + (b + 1)^2 + (c  + 1)^2 = 0 + 0+ 0

By comparison

(a - 1)^2 = 0

(b + 1)^2 = 0

(c  + 1)^2 = 0

Solving for (a - 1)^2 = 0

Take square root of both sides

a - 1 = 0

Add 1 to both sides

a - 1 + 1 = 0 + 1

a = 1

Solving for (b + 1)^2 = 0

Take square root of both sides

b + 1 = 0

Subtract 1 from both sides

b + 1 - 1 = 0 - 1

b = -1

Solving for (c  + 1)^2 = 0

Take square root of both sides

c + 1 = 0

Subtract 1 from both sides

c + 1 - 1 = 0 - 1

c = -1

Substitute the values of a, b and c in 2a - 3b + 4c

2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)

2a - 3b + 4c = 2 +3  -4

2a - 3b + 4c = 1

7 0
3 years ago
Which of the following statements is false? 1.5 &gt; 1,4
ArbitrLikvidat [17]

Answer:

the first one

Step-by-step explanation:

5 0
3 years ago
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