$a(0) = 1 \: \: \: \: \: \: b(0) = 2 \: \: \: \: \: c(0) = 3 \\ a(1) = b(0) + c(0) = 2 + 3 = 5 \\ b(1) = a(0) + c(0) = 1 + 3 = 4 \\ c(1) = a(0) + b(0) = 1 + 2 = 3 \\ \\ a(2) = b(1) + c(1) = 4 + 3 = 7 \\ b(2) = a(1) + c(1) = 5 + 3 = 8 \\ c(2) = a(1) + b(1) = 5 + 4 = 9$

$a(3) = b(2) + c(2) = 8 + 9 = 17\\ b(3) = a(2) + c(2) =7 + 9 = 16 \\ c(3) = a(2) + b(2) = 7 + 8 = 15 \\ \\ a(4) = b(3) + c(3) = 16 + 15 = 31 \\ b(4) = a(3) + c(3) = 17 + 15 = 32 \\ c(4) = a(3) + b(3) = 17 + 16 = 33$

$a(5) = 32 + 33 = 65 \\ b(5) = 31 + 33 = 64 \\ c(5) =31 + 32 = 63 \\ \\ a(6) = 64 + 63 = 127 \\ b(6) = 65 + 63 = 128 \\ c(6) = 65 + 64 = 129 \\$

$a(7) = 128 + 129 = 257 \\ b(7) = 127 + 129 = 256 \\c (7) = 127 + 128 = 255 \\ \\ a(8) = 256 + 255 = 511 \\ b(8) = 257 + 255 = 512 \\ c(8) = 257 + 256 = 513$

$a(9) = 512 + 513 = 1025 \\ b(9) = 511 + 513 = 1024 \\ c(9) = 511 + 512 = 1023 \\ \\ a(10) = 1024 + 1023 = 2047 \\ b(10) = 1025 + 1023 = 2048 \\ c(10) = 1025 + 1024 = 2049$

$a(n) + b(n) + c(n) = \\ 2(a(n - 1) + b(n - 1) + c(n - 1)) \\ 6 \times 2 {}^{n }$

$6 \times 2 {}^{n} > 100 \: 000 \\ 2 {}^{n} > \frac{100 \: 000}{6} \\ n > log {}^{2} ( \frac{100 \: 000}{6} ) \\ n > 14.02468 \\ n = 15$

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5 0

2 years ago

Choose the correct ordered pair. <br><br> Help please

emmainna [20.7K]

We can solve this system of equations using substitution.

2y + 2x = 8

Solve for x.

2x = -2y + 8

Divide all terms by 2.

x = -y + 4

Take the equation x = -y + 4 and substitute it into the second system of equation.

2y - 4x = 2

2y - 4(-y + 4) = 2

Simplify by distributing.

2y + 4y - 16 = 2

Combine like terms.

6y - 16 = 2

6y = 18

y = 3

Now that you know the numerical value of y, substitute y with 3 in both equations to find x.

1) 2y + 2x = 8

2(3) + 2x = 8

6 + 2x = 8

2x = 2

x = 1

2) 2y - 4x = 2

2(3) - 4x = 2

6 - 4x = 2

-4x = -4

x = 1

The value of x is 1, and the value of y is 3.

Solution: B) (1, 3)

8 0

4 years ago

The faculty of a certain four-year college consists of 179 teachers. There are 663 first-year students. The student-to-faculty r

DiKsa [7]

The answer is B
Hope it helps.

6 0

3 years ago

What is the value of cos0 ?

emmainna [20.7K]

Answer:

$\frac{9}{10}$

Step-by-step explanation:

The Law of Cosines works for any triangle and is given by:

$c^2=a^2+b^2-2ab\cos \gamma$ , where $a$ , $b$ , and $c$ are the sides of the triangle and $\gamma$ is the angle opposite to $c$ .

In the given triangle, assign the variables:

$c\implies 10$
$a\implies 10$
$b\implies 18$

Substituting these variables, we have:

$10^2=10^2+18^2-2(10)(18)\cod \cos \gamma,\\100=100+324-360\cdot \cos \gamma,\\100=424-360\cos \gamma,\\-324=-360\cos\gamma,\\\cos \gamma =\frac{-324}{-360}=\boxed{\frac{9}{10}}$

4 0

3 years ago